CBSE Class 12 · 2006 · Indefinite IntegralsLong Answer

Evaluate :

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

Solution

Write

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

, where

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

. Then

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

The first integral is

\frac{1}{2} lo g ∣2 x^{2} + 4 x - 3∣

. For the second,

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

, giving a standard

\frac{1}{u ^{2} - a ^{2}}

form:

- \frac{1}{2} \int \frac{d x}{( x + 1 ) ^{2} - ( 5/2 ) ^{2}} = - \frac{1}{2 10} lo g \frac{2 x + 2 - 3}{2 x + 2 + 3} .

\frac{1}{2} lo g 2 x^{2} + 4 x - 3 - \frac{1}{2 10} lo g \frac{2 x + 2 - 3}{2 x + 2 + 3} + C

CBSE Class 12 Board · 2006 · Subjective