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Integral Calculus

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Question : 26 of 73

Marks: +1, -0

Value of

\int\limits_{0}^{1} \;\frac{dx}{(1+x)(2+x)}

is

$\frac{\log 4}{3}$
$\frac{\log 2}{3}$
$\frac{\log 2}{3}$
none

Solution:

Concept:

Partial fraction decomposition to integrate rational function.

Explanation:

Write

\frac{1}{(1+x)(2+x)} = \frac{A}{1+x} + \frac{B}{2+x}

.

Multiply:

1 = A(2+x) + B(1+x)

.

Let

x = -1

:

1 = A(2-1) = A \implies A = 1

.

Let

x = -2

:

1 = B(1-2) = -B \implies B = -1

.

Thus

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

.

Integrate:

\int_0^1 \left(\frac{1}{1+x} - \frac{1}{2+x}\right) dx = \left[\ln|1+x| - \ln|2+x|\right]_0^1

.

Evaluate:

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

.

The options are

\frac{\ln 4}{3}

,

\frac{\ln 2}{3}

, and none. So correct choice is D.

Answer:

\ln \frac{4}{3}

, i.e., option D.

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