# ∫ 5 ( X 2 + 1 ) ( X + 2 ) D X - Mathematics

Sum
$\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx$

#### Solution

We have,

$I = \int\frac{5 dx}{\left( x^2 + 1 \right) \left( x + 2 \right)}$

$\text{Let }\frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 1}$

$\Rightarrow \frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{A \left( x^2 + 1 \right) + \left( Bx + C \right) \left( x + 2 \right)}{\left( x + 2 \right) \left( x^2 + 1 \right)}$

$\Rightarrow 5 = A \left( x^2 + 1 \right) + B x^2 + 2Bx + Cx + 2C$

$\Rightarrow 5 = \left( A + B \right) x^2 + \left( 2B + C \right) x + \left( A + 2C \right)$

$\text{Equating coefficients of like terms}$

$A + B = 0 . . . . . \left( 1 \right)$

$2B + C = 0 . . . . . \left( 2 \right)$

$A + 2C = 5 . . . . . \left( 3 \right)$

$\text{Solving (1), (2) and (3), we get}$

$A = 1$

$B = - 1$

$C = 2$

$\therefore \frac{5}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \frac{1}{x + 2} + \left( \frac{- x + 2}{x^2 + 1} \right)$

$\Rightarrow \int\frac{5 dx}{\left( x + 2 \right) \left( x^2 + 1 \right)} = \int\frac{dx}{x + 2} - \int\frac{x dx}{x^2 + 1} + 2\int\frac{dx}{x^2 + 1}$

$\text{Let }x^2 + 1 = t$

$\Rightarrow 2xdx = dt$

$\Rightarrow x dx = \frac{dt}{2}$

$\therefore I = \int\frac{dx}{x + 2} - \frac{1}{2}\int\frac{dt}{t} + 2\int\frac{dx}{x^2 + 1^2}$

$= \log \left| x + 2 \right| - \frac{1}{2} \log \left| t \right| + 2 \tan^{- 1} x + C'$

$= \log \left| x + 2 \right| - \frac{1}{2} \log \left| x^2 + 1 \right| + 2 \tan^{- 1} x + C'$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 36 | Page 177