CBSE Class 12 · 2013 · Indefinite IntegralsLong Answer

Evaluate :

\int \frac{5 x - 2}{1 + 2 x + 3 x ^{2}} d x

Answer: $\frac{5}{6} lo g 1 + 2 x + 3 x^{2} - \frac{11}{3 2} tan^{- 1} (\frac{3 x + 1}{2}) + C$

Solution

Write

5 x - 2 = \frac{5}{6} (6 x + 2) - \frac{11}{3}

, where

6 x + 2 = \frac{d}{d x} (3 x^{2} + 2 x + 1)

. Then

\int \frac{5 x - 2}{3 x ^{2} + 2 x + 1} d x = \frac{5}{6} \int \frac{6 x + 2}{3 x ^{2} + 2 x + 1} d x - \frac{11}{3} \int \frac{d x}{3 x ^{2} + 2 x + 1} .

The first integral is

\frac{5}{6} lo g ∣3 x^{2} + 2 x + 1∣

. For the second,

3 x^{2} + 2 x + 1 = 3 [(x + \frac{1}{3})^{2} + \frac{2}{9}]

- \frac{11}{3} \cdot \frac{1}{3} \int \frac{d x}{( x + \frac{1}{3} ) ^{2} + \frac{2}{9}} = - \frac{11}{3 2} tan^{- 1} (\frac{3 x + 1}{2}) .

\frac{5}{6} lo g 1 + 2 x + 3 x^{2} - \frac{11}{3 2} tan^{- 1} (\frac{3 x + 1}{2}) + C

CBSE Class 12 Board · 2013 · Subjective