CBSE Class 12 · 2007 · Indefinite IntegralsVery Short Answer

Evaluate :

\int \frac{1 + x ^{2}}{1 + x ^{4}} d x

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

Solution

Divide numerator and denominator by

x^{2}

\int \frac{1 + \frac{1}{x ^{2}}}{x ^{2} + \frac{1}{x ^{2}}} d x .

Put

u = x - \frac{1}{x}

, so

d u = (1 + \frac{1}{x ^{2}}) d x

and

x^{2} + \frac{1}{x ^{2}} = u^{2} + 2

. Then

\int \frac{d u}{u ^{2} + 2} = \frac{1}{2} tan^{- 1} \frac{u}{2} + C .

Since

u = x - \frac{1}{x} = \frac{x ^{2} - 1}{x}

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

CBSE Class 12 Board · 2007 · Subjective