First slide

Methods of integration

Question

If $\int \frac{x}{x^{2} - 4 x + 8} d x = K \log (x^{2} - 4 x + 8) +$ $\tan^{- 1} (\frac{x - 2}{2}) + C$ then the value of $K$ is

Easy

A

$1 / 2$
B

1
C

2
D

none of these

Solution

Completing the square, we get
$x^{2} - 4 x + 8 = (x - 2)^{2} + 4$ so put $x - 2 = u$
$\begin{array}{l} \int \frac{x}{x^{2} - 4 x + 8} d x = \int \frac{u + 2}{u^{2} + 4} d u \\ = \frac{1}{2} \int \frac{2 u}{u^{2} + 4} d u + 2 \int \frac{d u}{u^{2} + 4} \\ = \frac{1}{2} \log (u^{2} + 4) + \tan^{- 1} \frac{u}{2} + C \\ = \frac{1}{2} \log (x^{2} - 4 x + 8) + \tan^{- 1} \frac{x - 2}{2} + C . \end{array}$

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