Overview

We will explore the fundamental integration rules used in calculus, covering essential concepts such as the constant rule, exponential rule, power rule, and trigonometric rules. Each rule is illustrated with practical examples, helping readers gain confidence in integrating various functions. This document also covers the special integral forms, definite integral properties, and commonly used JEE results that are essential for competitive mathematics.

1. Integration Rules

1.1 Constant Rule

The constant rule states that the integral of a constant aa with respect to xx is the constant times xx plus the constant of integration CC:

∫a dx=ax+C\int a\,dx = ax + C

Example: Evaluate ∫4 dx\displaystyle\int 4\,dx

∫4 dx=4x+C\int 4\,dx = 4x + C

1.2 Exponential Rule

The exponential rule states that the integral of exe^x with respect to xx is exe^x plus the constant of integration CC:

∫ex dx=ex+C\int e^x\,dx = e^x + C

For a general base a>0a > 0, a≠1a \neq 1:

∫ax dx=axln⁡a+C\int a^x\,dx = \frac{a^x}{\ln a} + C

1.3 Power Rule

The power rule states that the integral of xnx^n with respect to xx (where nn is any real number except −1-1) is xn+1n+1\dfrac{x^{n+1}}{n+1} plus the constant of integration CC:

∫xn dx=xn+1n+1+C,where n≠−1\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad \text{where } n \neq -1

Example: Evaluate ∫x2 dx\displaystyle\int x^2\,dx

∫x2 dx=x33+C\int x^2\,dx = \frac{x^3}{3} + C

1.4 Multiplication by Constant

The multiplication by constant rule states that if we have a constant cc multiplied by a function f(x)f(x), then the integral of this product is equal to cc times the integral of f(x)f(x):

∫cf(x) dx=c∫f(x) dx\int c f(x)\,dx = c \int f(x)\,dx

Example: Evaluate ∫5x2 dx\displaystyle\int 5x^2\,dx

∫5x2 dx=5∫x2 dx=5⋅x33+C=5x33+C\int 5x^2\,dx = 5 \int x^2\,dx = 5 \cdot \frac{x^3}{3} + C = \frac{5x^3}{3} + C

1.5 Sum Rule

The sum rule states that the integral of the sum of two functions f(x)f(x) and g(x)g(x) is equal to the sum of their individual integrals:

∫(f(x)+g(x)) dx=∫f(x) dx+∫g(x) dx\int (f(x) + g(x))\,dx = \int f(x)\,dx + \int g(x)\,dx

Example: Evaluate ∫(2x2+3x) dx\displaystyle\int (2x^2 + 3x)\,dx

∫(2x2+3x) dx=∫2x2 dx+∫3x dx=2x33+3x22+C\int (2x^2 + 3x)\,dx = \int 2x^2\,dx + \int 3x\,dx = \frac{2x^3}{3} + \frac{3x^2}{2} + C

1.6 Difference Rule

The difference rule states that the integral of the difference of two functions is equal to the difference of their individual integrals:

∫(f(x)−g(x)) dx=∫f(x) dx−∫g(x) dx\int (f(x) - g(x))\,dx = \int f(x)\,dx - \int g(x)\,dx

Example: Evaluate ∫(5x2−2x) dx\displaystyle\int (5x^2 - 2x)\,dx

∫(5x2−2x) dx=∫5x2 dx−∫2x dx=5x33−x2+C\int (5x^2 - 2x)\,dx = \int 5x^2\,dx - \int 2x\,dx = \frac{5x^3}{3} - x^2 + C

1.7 Reciprocal Rule

The reciprocal rule, also known as the "Natural Logarithm Rule," states that the integral of 1x\dfrac{1}{x} with respect to xx is the natural logarithm of the absolute value of xx plus CC:

∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C

Example: Evaluate ∫12x dx\displaystyle\int \frac{1}{2x}\,dx

∫12x dx=12∫1x dx=12ln⁡∣x∣+C\int \frac{1}{2x}\,dx = \frac{1}{2} \int \frac{1}{x}\,dx = \frac{1}{2} \ln|x| + C

1.8 Trigonometric Rules (in radians)

The trigonometric rules involve integrals of trigonometric functions. Here are the important ones (where CC is the constant of integration):

∫sin⁡(x) dx=−cos⁡(x)+C∫cos⁡(x) dx=sin⁡(x)+C\int \sin(x)\,dx = -\cos(x) + C \qquad \int \cos(x)\,dx = \sin(x) + C
∫sec⁡2(x) dx=tan⁡(x)+C∫csc⁡2(x) dx=−cot⁡(x)+C\int \sec^2(x)\,dx = \tan(x) + C \qquad \int \csc^2(x)\,dx = -\cot(x) + C
∫sec⁡(x)tan⁡(x) dx=sec⁡(x)+C∫csc⁡(x)cot⁡(x) dx=−csc⁡(x)+C\int \sec(x)\tan(x)\,dx = \sec(x) + C \qquad \int \csc(x)\cot(x)\,dx = -\csc(x) + C
∫tan⁡(x) dx=ln⁡∣sec⁡(x)∣+C∫cot⁡(x) dx=ln⁡∣sin⁡(x)∣+C\int \tan(x)\,dx = \ln|\sec(x)| + C \qquad \int \cot(x)\,dx = \ln|\sin(x)| + C
∫sec⁡(x) dx=ln⁡∣sec⁡(x)+tan⁡(x)∣+C∫csc⁡(x) dx=ln⁡∣csc⁡(x)−cot⁡(x)∣+C\int \sec(x)\,dx = \ln|\sec(x) + \tan(x)| + C \qquad \int \csc(x)\,dx = \ln|\csc(x) - \cot(x)| + C

1.9 Integrals of Inverse Trigonometric Functions

∫11−x2 dx=arcsin⁡(x)+C∫(−11−x2)dx=arccos⁡(x)+C\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin(x) + C \qquad \int \left(-\frac{1}{\sqrt{1-x^2}}\right)dx = \arccos(x) + C
∫11+x2 dx=arctan⁡(x)+C∫(−11+x2)dx=arccot(x)+C\int \frac{1}{1+x^2}\,dx = \arctan(x) + C \qquad \int \left(-\frac{1}{1+x^2}\right)dx = \text{arccot}(x) + C
∫1∣x∣x2−1 dx=arcsec(x)+C∫(−1∣x∣x2−1)dx=arccsc(x)+C\int \frac{1}{|x|\sqrt{x^2-1}}\,dx = \text{arcsec}(x) + C \qquad \int \left(-\frac{1}{|x|\sqrt{x^2-1}}\right)dx = \text{arccsc}(x) + C

2. Special Integrals

These five forms appear constantly in JEE problems and should be memorized. For integrals of the form ∫dxax2+bx+c\displaystyle\int \frac{dx}{ax^2 + bx + c}, complete the square first to convert to one of these standard forms.

Type 1

∫dxx2+a2=1atan⁡−1xa+C\boxed{\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C}

Example: ∫dxx2+9=13tan⁡−1x3+C\displaystyle\int \frac{dx}{x^2 + 9} = \frac{1}{3}\tan^{-1}\frac{x}{3} + C

Type 2

∫dxx2−a2=12aln⁡∣x−ax+a∣+C\boxed{\int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C}

Example: ∫dxx2−4=14ln⁡∣x−2x+2∣+C\displaystyle\int \frac{dx}{x^2 - 4} = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Type 3

∫dxa2−x2=sin⁡−1xa+C\boxed{\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C}

Example: ∫dx25−x2=sin⁡−1x5+C\displaystyle\int \frac{dx}{\sqrt{25 - x^2}} = \sin^{-1}\frac{x}{5} + C

Type 4

∫dxx2+a2=ln⁡∣x+x2+a2∣+C\boxed{\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C}

Example: ∫dxx2+16=ln⁡∣x+x2+16∣+C\displaystyle\int \frac{dx}{\sqrt{x^2 + 16}} = \ln\left|x + \sqrt{x^2 + 16}\right| + C

Type 5

∫dxx2−a2=ln⁡∣x+x2−a2∣+C\boxed{\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left|x + \sqrt{x^2 - a^2}\right| + C}

Example: ∫dxx2−9=ln⁡∣x+x2−9∣+C\displaystyle\int \frac{dx}{\sqrt{x^2 - 9}} = \ln\left|x + \sqrt{x^2 - 9}\right| + C

3. Definite Integral Properties

A definite integral ∫abf(x) dx\displaystyle\int_a^b f(x)\,dx gives a number — the signed area under f(x)f(x) from aa to bb. The following properties allow many definite integrals to be evaluated without any computation at all.

Property 1 — Reversing Limits

Swapping the limits of integration changes the sign:

∫abf(x) dx=−∫baf(x) dx\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx

Property 2 — Zero-Width Interval

Integrating from a point to itself gives zero:

∫aaf(x) dx=0\int_a^a f(x)\,dx = 0

Property 3 — Additivity (Splitting the Interval)

An integral over [a,c][a, c] can be split at any intermediate point bb:

∫acf(x) dx=∫abf(x) dx+∫bcf(x) dx\int_a^c f(x)\,dx = \int_a^b f(x)\,dx + \int_b^c f(x)\,dx

This is especially useful when f(x)f(x) is defined differently on different sub-intervals (e.g., absolute value functions).

Property 4 — Linearity

Constants factor out and integrals of sums split, exactly as with indefinite integrals:

∫ab[k f(x)+g(x)] dx=k∫abf(x) dx+∫abg(x) dx\int_a^b [k\,f(x) + g(x)]\,dx = k\int_a^b f(x)\,dx + \int_a^b g(x)\,dx

Property 5 — Symmetry Property

∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx

Proof: Substitute u=a−xu = a - x. This is a special case of the King Property below.

Example: This immediately shows ∫0π/2sin⁡x dx=∫0π/2cos⁡x dx\displaystyle\int_0^{\pi/2} \sin x\,dx = \int_0^{\pi/2} \cos x\,dx.

Property 6 — Even and Odd Functions

For symmetric limits [−a,a][-a, a], always check whether ff is even or odd before computing.

For an even function where f(−x)=f(x)f(-x) = f(x):

∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx

For an odd function where f(−x)=−f(x)f(-x) = -f(x):

∫−aaf(x) dx=0\int_{-a}^a f(x)\,dx = 0

Example: ∫−22(x3+x5) dx=0\displaystyle\int_{-2}^{2} (x^3 + x^5)\,dx = 0 immediately, since x3+x5x^3 + x^5 is an odd function.

Example: ∫−11x4 dx=2∫01x4 dx=2⋅15=25\displaystyle\int_{-1}^{1} x^4\,dx = 2\int_0^1 x^4\,dx = 2 \cdot \frac{1}{5} = \frac{2}{5}

Property 7 — Periodic Functions

If f(x)f(x) has period TT (i.e. f(x+T)=f(x)f(x + T) = f(x)), then:

∫0nTf(x) dx=n∫0Tf(x) dx\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx

More generally, the integral over any full period is the same regardless of the starting point:

∫aa+Tf(x) dx=∫0Tf(x) dx\int_a^{a+T} f(x)\,dx = \int_0^T f(x)\,dx

Example: ∫04π∣sin⁡x∣ dx\displaystyle\int_0^{4\pi} |\sin x|\,dx. Since ∣sin⁡x∣|\sin x| has period π\pi and ∫0πsin⁡x dx=2\displaystyle\int_0^{\pi} \sin x\,dx = 2, the answer is 4×2=84 \times 2 = 8.

Property 8 — The King Property

This is the most powerful definite integral property and is a favourite in JEE problems.

∫abf(x) dx=∫abf(a+b−x) dx\boxed{\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx}

Proof: Substitute u=a+b−xu = a + b - x, so du=−dxdu = -dx. The limits swap from a→ba \to b to b→ab \to a, and flipping them back restores the sign.

The standard trick: When the integrand is not symmetric on its own, write II using the King Property to get a second expression for II, then add I+I=2II + I = 2I and simplify.

Example: Evaluate I=∫0π/2sin⁡xsin⁡x+cos⁡x dx\displaystyle I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx

Applying the King Property with a=0a = 0, b=π/2b = \pi/2:

I=∫0π/2cos⁡xcos⁡x+sin⁡x dxI = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}\,dx

Adding:

2I=∫0π/2sin⁡x+cos⁡xsin⁡x+cos⁡x dx=∫0π/21 dx=π2  ⟹  I=π42I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}

4. Commonly Used JEE Special Results

4.1 Wallis's Formula

A fast way to evaluate ∫0π/2sin⁡nx dx\displaystyle\int_0^{\pi/2} \sin^n x\,dx and ∫0π/2cos⁡nx dx\displaystyle\int_0^{\pi/2} \cos^n x\,dx (both are equal by the Symmetry Property):

∫0π/2sin⁡nx dx=∫0π/2cos⁡nx dx\int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx

For even powers (n=2mn = 2m):

∫0π/2sin⁡2mx dx=(2m−1)(2m−3)⋯3⋅1(2m)(2m−2)⋯4⋅2⋅π2\int_0^{\pi/2} \sin^{2m} x\,dx = \frac{(2m-1)(2m-3)\cdots 3 \cdot 1}{(2m)(2m-2)\cdots 4 \cdot 2} \cdot \frac{\pi}{2}

For odd powers (n=2m+1n = 2m+1):

∫0π/2sin⁡2m+1x dx=(2m)(2m−2)⋯4⋅2(2m+1)(2m−1)⋯3⋅1\int_0^{\pi/2} \sin^{2m+1} x\,dx = \frac{(2m)(2m-2)\cdots 4 \cdot 2}{(2m+1)(2m-1)\cdots 3 \cdot 1}

Quick examples:

∫0π/2sin⁡4x dx=3⋅14⋅2⋅π2=3π16\int_0^{\pi/2} \sin^4 x\,dx = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}
∫0π/2sin⁡5x dx=4⋅25⋅3⋅1=815\int_0^{\pi/2} \sin^5 x\,dx = \frac{4 \cdot 2}{5 \cdot 3 \cdot 1} = \frac{8}{15}

4.2 The ex[f(x)+f′(x)]e^x[f(x) + f'(x)] Result

When the integrand is an exponential multiplied by a function plus its derivative, the integral collapses neatly:

∫ex[f(x)+f′(x)] dx=exf(x)+C\boxed{\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C}

Why it works: Expanding ddx[exf(x)]=exf(x)+exf′(x)=ex[f(x)+f′(x)]\frac{d}{dx}[e^x f(x)] = e^x f(x) + e^x f'(x) = e^x[f(x) + f'(x)], so the integrand is already a derivative.

Example: ∫ex(tan⁡x+sec⁡2x) dx=extan⁡x+C\displaystyle\int e^x(\tan x + \sec^2 x)\,dx = e^x \tan x + C because f(x)=tan⁡xf(x) = \tan x and f′(x)=sec⁡2xf'(x) = \sec^2 x.

Example: ∫ex ⁣(1x+ln⁡x)dx=exln⁡x+C\displaystyle\int e^x\!\left(\frac{1}{x} + \ln x\right)dx = e^x \ln x + C because f(x)=ln⁡xf(x) = \ln x and f′(x)=1xf'(x) = \frac{1}{x}.

4.3 Integrals of eaxsin⁡bxe^{ax}\sin bx and eaxcos⁡bxe^{ax}\cos bx

Apply integration by parts twice and solve the resulting equation algebraically:

∫eaxsin⁡bx dx=eax(asin⁡bx−bcos⁡bx)a2+b2+C\int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2} + C
∫eaxcos⁡bx dx=eax(acos⁡bx+bsin⁡bx)a2+b2+C\int e^{ax}\cos bx\,dx = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2 + b^2} + C

4.4 Leibniz Rule (Differentiation Under the Integral Sign)

If the limits of a definite integral are functions of a parameter, differentiate using:

ddα∫g(α)h(α)f(x) dx=f(h(α))⋅h′(α)−f(g(α))⋅g′(α)\frac{d}{d\alpha}\int_{g(\alpha)}^{h(\alpha)} f(x)\,dx = f(h(\alpha))\cdot h'(\alpha) - f(g(\alpha))\cdot g'(\alpha)

Example: ddx∫0x2sin⁡t dt=sin⁡(x2)⋅2x=2xsin⁡(x2)\displaystyle\frac{d}{dx}\int_0^{x^2} \sin t\,dt = \sin(x^2) \cdot 2x = 2x\sin(x^2)

4.5 The General Symmetry Result

This generalises Property 5. For any power nn (including large odd integers like 20232023):

∫0π/2sin⁡nxsin⁡nx+cos⁡nx dx=π4\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\,dx = \frac{\pi}{4}

Proof: Apply the Symmetry Property to get a second expression, add, and cancel.

4.6 Riemann Sum as a Definite Integral

JEE Advanced frequently asks you to recognise limits of sums as definite integrals:

lim⁡n→∞1n∑r=1nf ⁣(rn)=∫01f(x) dx\lim_{n \to \infty} \frac{1}{n}\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx

Example: lim⁡n→∞1n∑r=1nrn=∫01x dx=23\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n \sqrt{\frac{r}{n}} = \int_0^1 \sqrt{x}\,dx = \frac{2}{3}

4.7 Inequality Property

If m≤f(x)≤Mm \leq f(x) \leq M for all x∈[a,b]x \in [a, b], then:

m(b−a)≤∫abf(x) dx≤M(b−a)m(b - a) \leq \int_a^b f(x)\,dx \leq M(b - a)

This is used to bound definite integrals without evaluating them.

5. Examples (Original)

Example 1

Evaluate ∫(4x6−2x3+7x−4) dx\displaystyle\int (4x^6 - 2x^3 + 7x - 4)\,dx

Solution: Apply the power rule to each term.

∫(4x6−2x3+7x−4) dx=47x7−12x4+72x2−4x+C\int (4x^6 - 2x^3 + 7x - 4)\,dx = \frac{4}{7}x^7 - \frac{1}{2}x^4 + \frac{7}{2}x^2 - 4x + C

Example 2

Evaluate ∫(z7−48z11−5z16) dz\displaystyle\int (z^7 - 48z^{11} - 5z^{16})\,dz

Solution: Apply the power rule to each term.

∫(z7−48z11−5z16) dz=18z8−4z12−517z17+C\int (z^7 - 48z^{11} - 5z^{16})\,dz = \frac{1}{8}z^8 - 4z^{12} - \frac{5}{17}z^{17} + C

Example 3

Evaluate ∫(10t−3+12t−9+4t3) dt\displaystyle\int (10t^{-3} + 12t^{-9} + 4t^3)\,dt

Solution: Apply the power rule to each term.

∫(10t−3+12t−9+4t3) dt=−102t2−128t8+44t4+C=−5t2−32t8+t4+C\int (10t^{-3} + 12t^{-9} + 4t^3)\,dt = -\frac{10}{2t^2} - \frac{12}{8t^8} + \frac{4}{4}t^4 + C = -\frac{5}{t^2} - \frac{3}{2t^8} + t^4 + C

Example 4

Evaluate ∫(t2−1)(4+3t) dt\displaystyle\int (t^2 - 1)(4 + 3t)\,dt

Solution: Use the distributive property, then apply the power rule.

∫(t2−1)(4+3t) dt=∫(3t3+4t2−3t−4) dt=34t4+43t3−32t2−4t+C\int (t^2 - 1)(4 + 3t)\,dt = \int (3t^3 + 4t^2 - 3t - 4)\,dt = \frac{3}{4}t^4 + \frac{4}{3}t^3 - \frac{3}{2}t^2 - 4t + C

Example 5

Evaluate ∫(sin⁡(x)+10csc⁡2(x)) dx\displaystyle\int (\sin(x) + 10\csc^2(x))\,dx

Solution: Use the integral of sin⁡(x)=−cos⁡(x)\sin(x) = -\cos(x) and csc⁡2(x)=−cot⁡(x)\csc^2(x) = -\cot(x).

∫(sin⁡(x)+10csc⁡2(x)) dx=−cos⁡(x)−10cot⁡(x)+C\int (\sin(x) + 10\csc^2(x))\,dx = -\cos(x) - 10\cot(x) + C

Example 6

Evaluate ∫(2cos⁡(w)−sec⁡(w)tan⁡(w)) dw\displaystyle\int (2\cos(w) - \sec(w)\tan(w))\,dw

Solution: Use the integral of cos⁡(w)=sin⁡(w)\cos(w) = \sin(w) and sec⁡(w)tan⁡(w)=sec⁡(w)\sec(w)\tan(w) = \sec(w).

∫(2cos⁡(w)−sec⁡(w)tan⁡(w)) dw=2sin⁡(w)−sec⁡(w)+C\int (2\cos(w) - \sec(w)\tan(w))\,dw = 2\sin(w) - \sec(w) + C

Example 7

Evaluate ∫(t3−e−t−4e−t)dt\displaystyle\int \left(t^3 - \frac{e^{-t} - 4}{e^{-t}}\right)dt

Solution: First simplify the expression inside the integral.

∫(t3−e−t−4e−t)dt=∫(t3−e−te−t+4e−t)dt=∫(t3−1+4et) dt\int \left(t^3 - \frac{e^{-t} - 4}{e^{-t}}\right)dt = \int \left(t^3 - \frac{e^{-t}}{e^{-t}} + \frac{4}{e^{-t}}\right)dt = \int (t^3 - 1 + 4e^t)\,dt

Now integrate each term:

=t44−t+4et+C= \frac{t^4}{4} - t + 4e^t + C

Example 8

Evaluate ∫(11+x2+121−x2)dx\displaystyle\int \left(\frac{1}{1+x^2} + \frac{12}{\sqrt{1-x^2}}\right)dx

Solution: Split into two separate integrals.

∫11+x2 dx+∫121−x2 dx=arctan⁡(x)+12arcsin⁡(x)+C\int \frac{1}{1+x^2}\,dx + \int \frac{12}{\sqrt{1-x^2}}\,dx = \arctan(x) + 12\arcsin(x) + C

Example 9

Evaluate ∫(x7−7x56+17x103)dx\displaystyle\int \left(\sqrt{x^7} - 7\sqrt[6]{x^5} + 17\sqrt[3]{x^{10}}\right)dx

Solution: Rewrite with fractional exponents, then apply the power rule.

∫(x7/2−7x5/6+17x10/3)dx=29x9/2−4211x11/6+5113x13/3+C\int \left(x^{7/2} - 7x^{5/6} + 17x^{10/3}\right)dx = \frac{2}{9}x^{9/2} - \frac{42}{11}x^{11/6} + \frac{51}{13}x^{13/3} + C
StatusSourceProblem NameDifficultyTags
AIMEVery Hard
Show TagsCombinatorics, Recursion

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