First slide

Methods of integration

Question

$\int \frac{dx}{(x + 2) (x^{2} + 1)}$ is equal to

Easy

A

$\frac{1}{5} \log | x + 2 | - \frac{1}{10} \log (x^{2} + 1) + \frac{2}{5} \tan^{- 1} x + C$
B

$\log | x + 2 | + \frac{1}{10} \log (x^{2} + 1) + \frac{1}{5} \tan^{- 1} x + C$
C

$5 \log | x + 2 | + 10 \log (x^{2} + 1) + \tan^{- 1} x + C$
D

None of the above

Solution

Let $\frac{1}{(x + 2) (x^{2} + 1)} = \frac{A}{x + 2} + \frac{Bx + C}{(x^{2} + 1)}$
$\Rightarrow 1 = A (x^{2} + 1) + (Bx + C) (x + 2)$
Put $x = - 2, we get A = \frac{1}{5}$
Now, comparing the coefficients of x² and constant term,
we get
$0 = A + B$
and $1 = A + 2 C$
$\Rightarrow B = - \frac{1}{5}, C = \frac{2}{5}$
So,
$\begin{array}{l} I = \frac{1}{5} \int \frac{dx}{x + 2} - \frac{1}{5} \int \frac{x}{x^{2} + 1} dx + \frac{2}{5} \int \frac{dx}{x^{2} + 1} \\ = \frac{1}{5} \log | x + 2 | - \frac{1}{10} \log (x^{2} + 1) + \frac{2}{5} \tan^{- 1} x + C \end{array}$

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