CBSE Class 12 · 2006 · Indefinite IntegralsLong Answer

Evaluate :

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

Answer: $\frac{3}{4} lo g 2 x^{2} - 2 x + 3 + \frac{5}{2} tan^{- 1} (\frac{2 x - 1}{5}) + C$

Solution

Write

3 x + 1 = \frac{3}{4} (4 x - 2) + \frac{5}{2}

, where

4 x - 2 = \frac{d}{d x} (2 x^{2} - 2 x + 3)

. Then

\int \frac{3 x + 1}{2 x ^{2} - 2 x + 3} d x = \frac{3}{4} \int \frac{4 x - 2}{2 x ^{2} - 2 x + 3} d x + \frac{5}{2} \int \frac{d x}{2 x ^{2} - 2 x + 3} .

The first integral is

\frac{3}{4} lo g ∣2 x^{2} - 2 x + 3∣

. For the second,

2 x^{2} - 2 x + 3 = 2 [(x - \frac{1}{2})^{2} + \frac{5}{4}]

\frac{5}{2} \cdot \frac{1}{2} \int \frac{d x}{( x - \frac{1}{2} ) ^{2} + \frac{5}{4}} = \frac{5}{2} tan^{- 1} (\frac{2 x - 1}{5}) .

\frac{3}{4} lo g 2 x^{2} - 2 x + 3 + \frac{5}{2} tan^{- 1} (\frac{2 x - 1}{5}) + C

CBSE Class 12 Board · 2006 · Subjective