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Indefinite-integration Formula Sheet

This page will help you to revise formulas and concepts of Indefinite-integration instantly for various exams.
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Indefinite integration is the process in calculus of finding a general antiderivative (or indefinite integral) of a function, without specifying the limits of integration, typically denoted by the symbol ∫.

Neetesh Kumar | June 02, 2024                                       \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space Share this Page on: Reddit icon Discord icon Email icon WhatsApp icon Telegram icon

1. Introduction to Indefinite Integration

If ff & FF are functions of xx such that F(x)=f(x)F' (x) = f(x) then the function FF is called a Primitive or Antiderivative or Integral of f(x)f(x) w.r.t. xx and is written symbolically as:
f(x)dx=F(x)+c    ddx(F(x)+c)=f(x)\int f(x) \, dx = F(x) + c \iff \frac{d}{dx}(F(x) + c) = f(x) where cc is called the constant of integration.

Note:

If f(x)dx=F(x)+c\int f(x) \, dx = F(x) + c, then f(ax+b)dx=F(ax+b)a+c,a0\int f(ax + b) \, dx = \frac{F(ax + b)}{a} + c, \, a \ne 0

2. Standard Results

(i) (ax+b)ndx=(ax+b)n+1a(n+1)+c;n1\int (ax + b)^n \, dx = \frac{(ax + b)^{n+1}}{a(n+1)} + c; \, n \ne -1

(ii) 1ax+bdx=1alnax+b+c\int \frac{1}{ax + b} \, dx = \frac{1}{a} \ln |ax + b| + c

(iii) eax+bdx=1aeax+b+c\int e^{ax + b} \, dx = \frac{1}{a} e^{ax + b} + c

(iv) dxacos2(ax+b)=1atan(ax+b)+c\int \frac{dx}{a \cos^2(ax + b)} = \frac{1}{a} \tan(ax + b) + c

(v) sin(ax+b)dx=1acos(ax+b)+c\int \sin(ax + b) \, dx = -\frac{1}{a} \cos(ax + b) + c

(vi) cos(ax+b)dx=1asin(ax+b)+c\int \cos(ax + b) \, dx = \frac{1}{a} \sin(ax + b) + c

(vii) tan(ax+b)dx=1alncos(ax+b)+c\int \tan(ax + b) \, dx = -\frac{1}{a} \ln |\cos(ax + b)| + c

(viii) cot(ax+b)dx=1alnsin(ax+b)+c\int \cot(ax + b) \, dx = \frac{1}{a} \ln |\sin(ax + b)| + c

(ix) sec2(ax+b)dx=1atan(ax+b)+c\int \sec^2(ax + b) \, dx = \frac{1}{a} \tan(ax + b) + c

(x) csc2(ax+b)dx=1acot(ax+b)+c\int \csc^2(ax + b) \, dx = -\frac{1}{a} \cot(ax + b) + c

(xi) csc(ax+b)cot(ax+b)dx=1acsc(ax+b)+c\int \csc(ax + b) \cot(ax + b) \, dx = -\frac{1}{a} \csc(ax + b) + c

(xii) sec(ax+b)tan(ax+b)dx=1asec(ax+b)+c\int \sec(ax + b) \tan(ax + b) \, dx = \frac{1}{a} \sec(ax + b) + c

(xiii) secxdx=lnsecx+tanx+c\int \sec x \, dx = \ln |\sec x + \tan x| + c

(xiv) cscxdx=lncscx+cotx+c\int \csc x \, dx = -\ln |\csc x + \cot x| + c

(xv) dxa2x2=sin1xa+c\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \frac{x}{a} + c

(xvi) dxa2+x2=sinh1xa+c\int \frac{dx}{\sqrt{a^2 + x^2}} = \sinh^{-1} \frac{x}{a} + c

(xvii) dxx2a2=cosh1xa+c\int \frac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1} \frac{x}{a} + c

(xviii) dxx2a2=12alnxax+a+c\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + c

(xix) dxx2+a2=1atan1xa+c\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c

(xx) dxa2x2=12alna+xax+c\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + c

(xxi) dxx2+a2=1atan1xa+c\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + c

(xxii) dx(x2a2)2=x2a2(x2a2)+c\int \frac{dx}{(x^2 - a^2)^2} = \frac{x}{2a^2(x^2 - a^2)} + c

(xxiii) xdxx2a2=12lnx2a2+c\int \frac{x \, dx}{x^2 - a^2} = \frac{1}{2} \ln |x^2 - a^2| + c

(xxiv) xdxx2+a2=12ln(x2+a2)+c\int \frac{x \, dx}{x^2 + a^2} = \frac{1}{2} \ln (x^2 + a^2) + c

(xxv) 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

(xxvi) 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

3. Techniques of Integration

(a) Substitution or change of independent variable:

Integral I=f(x)dxI = \int f(x) \, dx is changed to f(g(t))g(t)dt\int f(g(t)) g'(t) \, dt, by a suitable substitution x=g(t)x = g(t) provided the latter integral is easier to integrate.

Some standard substitutions:

  1. f(x)nf(x)dx\int f(x)^n f'(x) \, dx put f(x)=tf(x) = t & proceed.

  2. dxa2x2,dxa2+x2,dxx2a2\int \frac{dx}{\sqrt{a^2 - x^2}}, \int \frac{dx}{\sqrt{a^2 + x^2}}, \int \frac{dx}{\sqrt{x^2 - a^2}} Express ax2+bx+cax^2 + bx + c in the form of a perfect square & then apply the standard results.

  3. px+qax2+bx+cdx\int \frac{px + q}{ax^2 + bx + c} \, dx Express px+qpx + q as A×A \times (differential coefficient of the quadratic term of the denominator) + BB.

  4. ef(x)(f(x)+f(x))dx=ef(x)+c\int e^{f(x)}(f(x) + f'(x)) \, dx = e^{f(x)} + c

  5. (f(x)+xf(x))dx=xf(x)xdx\int (f(x) + x f'(x)) \, dx = x f(x) - \int x \, dx

  6. dxxn(xn+1)\int \frac{dx}{x^n (x^n + 1)}, take xnx^n common & put 1+xn=t1 + x^{-n} = t.

  7. dxx(x2+a2)\int \frac{dx}{x (x^2 + a^2)}, take x2x^2 common & put 1+x2=t1 + x^{-2} = t.

  8. dxx(1+xn)\int \frac{dx}{x (1 + x^n)}, take xnx^n common and put 1+xn=t1 + x^{-n} = t.

  9. dxa+bsin2x\int \frac{dx}{a+bsin^2x} OR dxa+bcos2x\int \frac{dx}{a+bcos^2x} OR dxasin2x+bsinxcosx+ccos2x\int \frac{dx}{asin^2x + bsinxcosx + ccos^2x}
    then multiply the Numerator and Denominator by Sec2^2x and Put tanx = t

  10. dxa+bsinx\int \frac{dx}{a+bsinx} OR dxa+bcosx\int \frac{dx}{a+bcosx} OR dxa+bsinx+ccosx\int \frac{dx}{a+bsinx+ccosx}
    Convert sines & cosines into their respective tangents of half the angles, put tanx2=t\frac{x}{2} = t

(b) Integration by parts:

u.vdx=uvdx[dudx.v.dx]dx\int u.v dx = u\int v dx - \int[\frac{du}{dx}.\int v.dx]dx

where uu & vv are differentiable functions.

Note:

While using integration by parts, choose uu & vv such that:

(i) vdx\int v \, dx & (ii) [dudx.vdx]dx\int[\frac{du}{dx}. v \, dx]dx is simple to integrate.

This is generally obtained by keeping the order of uu & vv as per the order of the letters in ILATE, where the I-Inverse function, L-Logarithmic function, A-Algebraic function, T-Trigonometric function & E-Exponential function.

(c) Partial fraction:

A rational function is defined as the ratio of two polynomials in the form P(x)Q(x)\frac{P(x)}{Q(x)}, where P(x)P(x) and Q(x)Q(x) are polynomials in xx and Q(x)0Q(x) \ne 0. If the degree of P(x)P(x) is less than that of Q(x)Q(x), then the rational function is called proper. Otherwise, it is called improper. The long division process can reduce the improper rational function to the proper rational functions. Thus, if P(x)Q(x)\frac{P(x)}{Q(x)} is improper, then P(x)Q(x)=T(x)+P1(x)Q(x)\frac{P(x)}{Q(x)} = T(x) + \frac{P_1(x)}{Q(x)}

where T(x)T(x) is a polynomial in xx and P1(x)Q(x)\frac{P_1(x)}{Q(x)} is a proper rational function. Writing the integrand as a sum of simpler rational functions by partial fraction decomposition is always possible. After this, the integration can be carried out easily using the already-known methods.

4. Various Forms of the Partial Fraction

(i) 2px+qx+r(xa)(xb)(xc)=Axa+Bxb+Cxc\frac{2px + qx + r}{(x - a) (x - b)(x - c)} = \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c}

(ii) 2px+qx+r(xa)2(xb)=Axa+B(xa)2+Cxb\frac{2px + qx + r}{(x - a)^2 (x - b)} = \frac{A}{x - a} + \frac{B}{(x - a)^2} + \frac{C}{x - b}

(iii) 2px+qx+r(xa)(x2+bx+c)=Axa+Bx+Cx2+bx+c\frac{2px + qx + r}{(x - a)(x^2 + bx + c)} = \frac{A}{x - a} + \frac{Bx + C}{x^2 + bx + c}

where x2+bx+cx^2 + bx + c cannot be factorized further

Note:
In competitive exams, partial fractions are generally found by inspection by noting the following fact:
1(xa)(xb)=1ab(1xa1xb)\frac{1}{(x - a)(x - b)} = \frac{1}{a - b} \left(\frac{1}{x - a} - \frac{1}{x - b}\right)

It can be applied when x2x^2 or any other function exists in all places of xx.

Examples:

  1. 1(x2+1)(x2+3)=12(1t+11t+3)\frac{1}{(x^2 + 1)(x^2 + 3)} = \frac{1}{2} \left(\frac{1}{t + 1} - \frac{1}{t + 3}\right) {take x2=tx^2 = t}

  2. 1x4+x2=1x2(x2+1)=12(1x21x2+1)\frac{1}{x^4 + x^2} = \frac{1}{x^2(x^2 + 1)} = \frac{1}{2} \left(\frac{1}{x^2} - \frac{1}{x^2 + 1}\right)

  3. xx3+x2=1x(x+1)=1x1x+1\frac{x}{x^3 + x^2} = \frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}

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