Evaluate : $\int \frac{2x^{2}+1}{x^{2}\left(x^{2}+4\right)}\,dx$

$-\dfrac{1}{4x}+\dfrac{7}{8}\tan^{-1}\!\left(\dfrac{x}{2}\right)+C$

CBSE Class 12 · 2013 · Indefinite IntegralsLong Answer

Evaluate :

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

Answer: $- \frac{1}{4 x} + \frac{7}{8} tan^{- 1} (\frac{x}{2}) + C$

Split the numerator:

2 x^{2} + 1 = 2 (x^{2} + 4) - 7

, so

\frac{2 x ^{2} + 1}{x ^{2} ( x ^{2} + 4 )} = \frac{2}{x ^{2}} - \frac{7}{x ^{2} ( x ^{2} + 4 )} .

Writing

\frac{1}{x ^{2} ( x ^{2} + 4 )} = \frac{1}{4} (\frac{1}{x ^{2}} - \frac{1}{x ^{2} + 4})

, the integrand reduces to

\frac{1}{4} \cdot \frac{1}{x ^{2}} + \frac{7}{4} \cdot \frac{1}{x ^{2} + 4} .

Integrating,

- \frac{1}{4 x} + \frac{7}{8} tan^{- 1} (\frac{x}{2}) + C

CBSE Class 12 Board · 2013 · Subjective