# Integration by Partial Fraction Formula

Here you will learn integration by partial fraction formula and integration of irrational functions.

Let’s begin –

## Integration by Partial Fraction Formula

(i) Integration of Rational Functions

S.No form of rational function form of partial fraction
1 $$px^2+qx+r\over {(x-a)(x-b)(x-c)}$$ $$A\over {x-a}$$ + $$B\over {x-b}$$ + $$C\over {x-c}$$
2 $$px^2+qx+r\over {{(x-a)}^2(x-b)}$$ $$A\over {x-a}$$ + $$B\over {(x-a)}^2$$ + $$C\over {x-b}$$
3 $$px^2+qx+r\over {(x-a)(x^2+bx+c)}$$ $$A\over {x-a}$$ + $$Bx+C\over {x^2+bx+c}$$

Example : Evaluate $$\int$$ $$x\over {(x-2)(x-5)}$$ dx

Solution : We have, $$\int$$ $$x\over {(x-2)(x-5)}$$ dx

Let $$x\over {(x-2)(x-5)}$$ = $$A\over {x-2}$$ + $$B\over {x-5}$$

or   x = A(x+5) + B(x-2)

by comparing the coefficients, we get

A = 2/7 and B = 5/7 so that

$$\int$$ $$x\over {(x-2)(x-5)}$$ dx = $$2\over 7$$ $$\int$$$$dx\over x-2$$ + $$5\over 7$$ $$\int$$$$dx\over x+5$$

= $$2\over 7$$ ln|x-2| + $$5\over 7$$ ln|x+5| + C

Example : Evaluate $$\int$$ $$2x\over {(x^2+1)(x^2+2)}$$ dx

Solution : Let I = $$\int$$ $$2x\over {(x^2+1)(x^2+2)}$$ dx

Putting $$x^2$$ = t and 2xdx = dt, we get

I = $$\int$$ $$dt\over {(t+1)(t+2)}$$

Let $$1\over {(t+1)(t+2)}$$ = $$A\over t+1$$ + $$B\over t+2$$ …….(i)

$$\implies$$ 1 = A(t+2) + B(t+1) ……..(ii)

Putting t = -2 in (ii), we obtain B = -1

Putting t = -1 in (ii), we obtain A = 1

Putting value of A and B in (i), we get

$$1\over {(t+1)(t+2)}$$ = $$1\over t+1$$ – $$1\over t+2$$

I = $$\int$$ $$1\over {(t+1)(t+2)}$$

$$\implies$$ I = $$\int$$ $$1\over t+1$$dt – $$\int$$ $$1\over t+2$$dt

$$\implies$$ I = log|t+1| – log|t+2| + C

$$log|x^2+1|$$ – $$log|x^2+2|$$ + C

(ii)  Integration of Irrational Functions

(a) $$\int$$ $$dx\over {(ax + b)\sqrt{px+q}}$$ & $$\int$$ $$dx\over {(ax^2 + bx + c)\sqrt{px+q}}$$; put px+q = $$t^2$$

(b)  $$\int$$ $$dx\over {(ax + b)\sqrt{px^2+qx+r}}$$; put ax+b = $$1\over t$$; $$\int$$ $$dx\over {(ax^2 + b)\sqrt{px^2+q}}$$; put x = $$1\over t$$