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CBSE Class 12 · 2017 · Indefinite IntegralsVery Short Answer

Find

\int \frac{d x}{x ^{2} + 4 x + 8}

Answer: $\frac{1}{2} tan^{- 1} (\frac{x + 2}{2}) + C$

Solution

Complete the square:

x^{2} + 4 x + 8 = (x + 2)^{2} + 2^{2}

. Using

\int \frac{d x}{u ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{u}{a}

with

u = x + 2

,

a = 2

:

\frac{1}{2} tan^{- 1} (\frac{x + 2}{2}) + C

CBSE Class 12 Board · 2017 · Subjective

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