Integrals of Some Particular Functions

There are six very important standard integrals that appear repeatedly in calculus. These should be memorized carefully because many complicated-looking integrals reduce to one of these forms after a little algebra, substitution, or completing the square.

Without Square Roots

  1. ∫dxx2−a2=12alog⁡∣x−ax+a∣+C\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C
  2. ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C
  3. ∫dxx2+a2=1atan⁡−1(xa)+C\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C

With Square Roots

  1. ∫dxx2−a2=log⁡∣x+x2−a2∣+C\int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C
  2. ∫dxa2−x2=sin⁡−1(xa)+C\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
  3. ∫dxx2+a2=log⁡∣x+x2+a2∣+C\int \frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C

These formulas are valid for a>0a>0. In actual problems, the algebraic expression in the denominator or under the square root is often first converted into one of these standard forms.

Integration by Completing the Square

To evaluate integrals like ∫dxax2+bx+c\int \frac{dx}{ax^2+bx+c} or ∫dxax2+bx+c,\int \frac{dx}{\sqrt{ax^2+bx+c}}, we usually rewrite the quadratic expression as a perfect square plus or minus a constant. This is done by completing the square.

General Method

  1. Factor out the coefficient of x2x^2 if necessary, so that the coefficient of x2x^2 inside the bracket becomes 1.
  2. Take half of the coefficient of xx, square it, and add and subtract the same quantity.
  3. Group the first three terms into a perfect square.
  4. Rewrite the integral so that it matches one of the standard forms above.

For example, x2+6x+13=(x+3)2+4x^2+6x+13=(x+3)^2+4 and 5−4x−x2=9−(x+2)2.5-4x-x^2=9-(x+2)^2.

After this transformation, the integral becomes much easier and can be evaluated using a standard formula.

Integrals of the Form Linear / Quadratic

Integrals of the type ∫px+qax2+bx+c dx\int \frac{px+q}{ax^2+bx+c}\,dx and ∫px+qax2+bx+c dx\int \frac{px+q}{\sqrt{ax^2+bx+c}}\,dx often cannot be solved directly by a single substitution because the numerator is not exactly the derivative of the quadratic denominator.

Working Rule

  1. Express the numerator as px+q=A⋅ddx(ax2+bx+c)+B.px+q=A\cdot \frac{d}{dx}(ax^2+bx+c)+B. Since ddx(ax2+bx+c)=2ax+b,\frac{d}{dx}(ax^2+bx+c)=2ax+b, we actually write px+q=A(2ax+b)+B.px+q=A(2ax+b)+B.
  2. Compare coefficients of xx and the constant term to find AA and BB.
  3. Split the integral into two parts:
  • one involving the derivative of the quadratic divided by the quadratic, which gives a logarithm,
  • the other involving only a constant divided by the quadratic, which is handled by completing the square.

This method is extremely useful and appears frequently in board and competitive exams.

Integration by Partial Fractions

A rational function is a quotient of two polynomials: P(x)Q(x),Q(x)≠0.\frac{P(x)}{Q(x)}, \qquad Q(x)\neq 0.

  • Proper Rational Function: degree of P(x)P(x) is less than degree of Q(x)Q(x).
  • Improper Rational Function: degree of P(x)P(x) is greater than or equal to degree of Q(x)Q(x).

Before using partial fractions, an improper rational function must first be converted into a polynomial plus a proper rational function by polynomial long division.

Once the fraction is proper, it can be decomposed into simpler fractions called partial fractions.

Standard Forms:

  1. Distinct linear factors: px+q(x−a)(x−b)=Ax−a+Bx−b\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}
  2. Repeated linear factor: px+q(x−a)2=Ax−a+B(x−a)2\frac{px+q}{(x-a)^2}=\frac{A}{x-a}+\frac{B}{(x-a)^2}
  3. Three distinct linear factors: px2+qx+r(x−a)(x−b)(x−c)=Ax−a+Bx−b+Cx−c\frac{px^2+qx+r}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}
  4. Repeated and distinct linear factors: px2+qx+r(x−a)2(x−b)=Ax−a+B(x−a)2+Cx−b\frac{px^2+qx+r}{(x-a)^2(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}
  5. Irreducible quadratic factor: px2+qx+r(x−a)(x2+bx+c)=Ax−a+Bx+Cx2+bx+c\frac{px^2+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 into real linear factors.

After decomposition, each partial fraction is integrated separately using logarithmic forms, power-rule forms, or standard quadratic integrals.

Solved Examples

Example 1: Direct Application of Standard Formula

Evaluate: ∫dxx2−16\int \frac{dx}{x^2-16}

Solution: Step 1: Recognize the denominator as a difference of squares: x2−16=x2−42.x^2-16=x^2-4^2. So this matches the standard form ∫dxx2−a2.\int \frac{dx}{x^2-a^2}. Here, a=4a=4.

Step 2: Use the formula ∫dxx2−a2=12alog⁡∣x−ax+a∣+C.\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C.

Step 3: Substitute a=4a=4: ∫dxx2−16=12⋅4log⁡∣x−4x+4∣+C.\int \frac{dx}{x^2-16}=\frac{1}{2\cdot 4}\log\left|\frac{x-4}{x+4}\right|+C.

Step 4: Simplify: =18log⁡∣x−4x+4∣+C.=\frac{1}{8}\log\left|\frac{x-4}{x+4}\right|+C.

Answer: 18log⁡∣x−4x+4∣+C\frac{1}{8}\log\left|\frac{x-4}{x+4}\right|+C

Example 2: Completing the Square (Without Root)

Evaluate: ∫dxx2−6x+13\int \frac{dx}{x^2-6x+13}

Solution: Step 1: Complete the square in the denominator. Take half of −6-6, which is −3-3, and square it to get 99. So, x2−6x+13=x2−6x+9+4=(x−3)2+4.x^2-6x+13=x^2-6x+9+4=(x-3)^2+4.

Step 2: Rewrite the integral: ∫dx(x−3)2+22.\int \frac{dx}{(x-3)^2+2^2}.

Step 3: Compare with the standard form ∫dxX2+a2=1atan⁡−1(Xa)+C.\int \frac{dx}{X^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{X}{a}\right)+C. Here, X=x−3X=x-3 and a=2a=2.

Step 4: Apply the formula: ∫dx(x−3)2+22=12tan⁡−1(x−32)+C.\int \frac{dx}{(x-3)^2+2^2}=\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right)+C.

Answer: 12tan⁡−1(x−32)+C\frac{1}{2}\tan^{-1}\left(\frac{x-3}{2}\right)+C

Example 3: Completing the Square (With Root and Negative x2x^2)

Evaluate: ∫dx5−4x−x2\int \frac{dx}{\sqrt{5-4x-x^2}}

Solution: Step 1: Rewrite the expression under the root by factoring out the negative sign from the quadratic part: 5−4x−x2=5−(x2+4x).5-4x-x^2=5-(x^2+4x).

Step 2: Complete the square inside the bracket: x2+4x=x2+4x+4−4=(x+2)2−4.x^2+4x=x^2+4x+4-4=(x+2)^2-4. So, 5−(x2+4x)=5−[(x+2)2−4]=9−(x+2)2.5-(x^2+4x)=5-\big[(x+2)^2-4\big]=9-(x+2)^2. Thus, 5−4x−x2=32−(x+2)2.5-4x-x^2=3^2-(x+2)^2.

Step 3: Rewrite the integral: ∫dx32−(x+2)2.\int \frac{dx}{\sqrt{3^2-(x+2)^2}}.

Step 4: Use the standard formula ∫dxa2−X2=sin⁡−1(Xa)+C.\int \frac{dx}{\sqrt{a^2-X^2}}=\sin^{-1}\left(\frac{X}{a}\right)+C. Here, a=3a=3 and X=x+2X=x+2.

Step 5: Therefore, ∫dx5−4x−x2=sin⁡−1(x+23)+C.\int \frac{dx}{\sqrt{5-4x-x^2}}=\sin^{-1}\left(\frac{x+2}{3}\right)+C.

Answer: sin⁡−1(x+23)+C\sin^{-1}\left(\frac{x+2}{3}\right)+C

Example 4: Linear / Quadratic Form

Evaluate: ∫x+22x2+6x+5dx\int \frac{x+2}{2x^2+6x+5}dx

Solution: Step 1: The denominator is 2x2+6x+5,2x^2+6x+5, whose derivative is 4x+6.4x+6. We express the numerator x+2x+2 in the form x+2=A(4x+6)+B.x+2=A(4x+6)+B.

Step 2: Compare coefficients. From the coefficient of xx: 1=4A⇒A=14.1=4A \Rightarrow A=\frac{1}{4}. From the constant term: 2=6A+B=6(14)+B=32+B,2=6A+B=6\left(\frac{1}{4}\right)+B=\frac{3}{2}+B, so B=12.B=\frac{1}{2}.

Step 3: Rewrite the integral: ∫x+22x2+6x+5dx=14∫4x+62x2+6x+5dx+12∫dx2x2+6x+5.\int \frac{x+2}{2x^2+6x+5}dx=\frac{1}{4}\int \frac{4x+6}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{dx}{2x^2+6x+5}.

Step 4: Evaluate the first integral. Let t=2x2+6x+5⇒dt=(4x+6)dx.t=2x^2+6x+5 \Rightarrow dt=(4x+6)dx. Then, ∫4x+62x2+6x+5dx=∫dtt=log⁡∣t∣=log⁡∣2x2+6x+5∣.\int \frac{4x+6}{2x^2+6x+5}dx=\int \frac{dt}{t}=\log|t|=\log|2x^2+6x+5|.

Step 5: Evaluate the second integral: I=∫dx2x2+6x+5.I=\int \frac{dx}{2x^2+6x+5}. Factor out 2: I=12∫dxx2+3x+52.I=\frac{1}{2}\int \frac{dx}{x^2+3x+\frac{5}{2}}. Complete the square: x2+3x+52=x2+3x+94−94+52=(x+32)2+14.x^2+3x+\frac{5}{2}=x^2+3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{2}=(x+\tfrac{3}{2})^2+\frac{1}{4}. So, I=12∫dx(x+32)2+(12)2.I=\frac{1}{2}\int \frac{dx}{(x+\tfrac{3}{2})^2+(\tfrac{1}{2})^2}. Using the standard formula, I=12⋅11/2tan⁡−1(x+3/21/2)=tan⁡−1(2x+3).I=\frac{1}{2}\cdot \frac{1}{1/2}\tan^{-1}\left(\frac{x+3/2}{1/2}\right)=\tan^{-1}(2x+3).

Step 6: Multiply by the outside coefficient 12\frac{1}{2} from Step 3 and combine everything: ∫x+22x2+6x+5dx=14log⁡∣2x2+6x+5∣+12tan⁡−1(2x+3)+C.\int \frac{x+2}{2x^2+6x+5}dx=\frac{1}{4}\log|2x^2+6x+5|+\frac{1}{2}\tan^{-1}(2x+3)+C.

Answer: 14log⁡∣2x2+6x+5∣+12tan⁡−1(2x+3)+C\frac{1}{4}\log|2x^2+6x+5|+\frac{1}{2}\tan^{-1}(2x+3)+C

Example 5: Basic Partial Fractions

Evaluate: ∫dx(x+1)(x+2)\int \frac{dx}{(x+1)(x+2)}

Solution: Step 1: Resolve into partial fractions: 1(x+1)(x+2)=Ax+1+Bx+2.\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}.

Step 2: Multiply both sides by (x+1)(x+2)(x+1)(x+2): 1=A(x+2)+B(x+1).1=A(x+2)+B(x+1).

Step 3: Find the constants. Put x=−1x=-1: 1=A(1)⇒A=1.1=A(1) \Rightarrow A=1. Put x=−2x=-2: 1=B(−1)⇒B=−1.1=B(-1) \Rightarrow B=-1.

Step 4: Rewrite the integral: ∫(1x+1−1x+2)dx.\int \left(\frac{1}{x+1}-\frac{1}{x+2}\right)dx.

Step 5: Integrate term by term: =log⁡∣x+1∣−log⁡∣x+2∣+C.=\log|x+1|-\log|x+2|+C.

Step 6: Combine the logarithms: =log⁡∣x+1x+2∣+C.=\log\left|\frac{x+1}{x+2}\right|+C.

Answer: log⁡∣x+1x+2∣+C\log\left|\frac{x+1}{x+2}\right|+C

Example 6: Improper Rational Function

Evaluate: ∫x2+1x2−5x+6dx\int \frac{x^2+1}{x^2-5x+6}dx

Solution: Step 1: Since the degree of the numerator equals the degree of the denominator, this is an improper rational function. Perform polynomial division: x2+1x2−5x+6=1+5x−5x2−5x+6.\frac{x^2+1}{x^2-5x+6}=1+\frac{5x-5}{x^2-5x+6}.

Step 2: Factor the denominator of the proper fraction: x2−5x+6=(x−2)(x−3).x^2-5x+6=(x-2)(x-3). So, 5x−5(x−2)(x−3)=Ax−2+Bx−3.\frac{5x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}.

Step 3: Multiply through by (x−2)(x−3)(x-2)(x-3): 5x−5=A(x−3)+B(x−2).5x-5=A(x-3)+B(x-2).

Step 4: Find AA and BB. Put x=2x=2: 10−5=A(−1)⇒A=−5.10-5=A(-1) \Rightarrow A=-5. Put x=3x=3: 15−5=B(1)⇒B=10.15-5=B(1) \Rightarrow B=10.

Step 5: Rewrite the integrand: x2+1x2−5x+6=1−5x−2+10x−3.\frac{x^2+1}{x^2-5x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}.

Step 6: Integrate term by term: ∫x2+1x2−5x+6dx=∫1 dx−5∫dxx−2+10∫dxx−3.\int \frac{x^2+1}{x^2-5x+6}dx=\int 1\,dx-5\int \frac{dx}{x-2}+10\int \frac{dx}{x-3}. Thus, =x−5log⁡∣x−2∣+10log⁡∣x−3∣+C.=x-5\log|x-2|+10\log|x-3|+C.

Answer: x−5log⁡∣x−2∣+10log⁡∣x−3∣+Cx-5\log|x-2|+10\log|x-3|+C

Example 7: Repeated Linear Factors

Evaluate: ∫3x−2(x+1)2(x+3)dx\int \frac{3x-2}{(x+1)^2(x+3)}dx

Solution: Step 1: Since (x+1)2(x+1)^2 is a repeated linear factor, write 3x−2(x+1)2(x+3)=Ax+1+B(x+1)2+Cx+3.\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}.

Step 2: Multiply both sides by (x+1)2(x+3)(x+1)^2(x+3): 3x−2=A(x+1)(x+3)+B(x+3)+C(x+1)2.3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)^2.

Step 3: Find BB and CC by substitution. Put x=−1x=-1: 3(−1)−2=B(2)⇒−5=2B⇒B=−52.3(-1)-2=B(2) \Rightarrow -5=2B \Rightarrow B=-\frac{5}{2}. Put x=−3x=-3: 3(−3)−2=C(−2)2⇒−11=4C⇒C=−114.3(-3)-2=C(-2)^2 \Rightarrow -11=4C \Rightarrow C=-\frac{11}{4}.

Step 4: Find AA. Compare coefficients of x2x^2 on both sides. Since the left side has no x2x^2 term, A+C=0⇒A=−C=114.A+C=0 \Rightarrow A=-C=\frac{11}{4}.

Step 5: Rewrite the integral: ∫(11/4x+1−5/2(x+1)2−11/4x+3)dx.\int \left(\frac{11/4}{x+1}-\frac{5/2}{(x+1)^2}-\frac{11/4}{x+3}\right)dx.

Step 6: Integrate each term: ∫11/4x+1dx=114log⁡∣x+1∣,\int \frac{11/4}{x+1}dx=\frac{11}{4}\log|x+1|, ∫−5/2(x+1)2dx=−52∫(x+1)−2dx=52(x+1),\int -\frac{5/2}{(x+1)^2}dx=-\frac{5}{2}\int (x+1)^{-2}dx=\frac{5}{2(x+1)}, ∫−11/4x+3dx=−114log⁡∣x+3∣.\int -\frac{11/4}{x+3}dx=-\frac{11}{4}\log|x+3|.

Step 7: Combine: =114log⁡∣x+1∣+52(x+1)−114log⁡∣x+3∣+C.=\frac{11}{4}\log|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\log|x+3|+C. Using logarithm properties, =114log⁡∣x+1x+3∣+52(x+1)+C.=\frac{11}{4}\log\left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C.

Answer: 114log⁡∣x+1x+3∣+52(x+1)+C\frac{11}{4}\log\left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C