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MHTCET 10 May 2024 Shift 2 Paper

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Question : 121 of 150

Marks: +1, -0

\int_{0}^{a} \frac{x-a}{x+a} d x=

$a-2 a \log 2$
$a-a \log 2$
$a+2 a \log 2$
$a+a \log 2$

Solution:

Concept:

Use substitution to simplify the integrand, then apply standard logarithmic integration.

Explanation:

Let

I = \int_{0}^{a} \frac{x-a}{x+a} dx

.

Put

x+a = t

, so

x = t-a

and

dx = dt

.

When

x=0

,

t=a

; when

x=a

,

t=2a

.

The integrand becomes

\frac{t-a-a}{t} = \frac{t-2a}{t}

.

Therefore,

I = \int_{a}^{2a} \frac{t-2a}{t} dt

.

Split the integral:

I = \int_{a}^{2a} 1\,dt - 2a\int_{a}^{2a} \frac{1}{t}dt

.

Evaluate:

I = [t]_{a}^{2a} - 2a[\log t]_{a}^{2a}

.

This gives

I = a - 2a(\log 2a - \log a)

.

Using logarithm properties,

I = a - 2a\log 2

.

Answer:

I = a - 2a \log 2

, which is option A.

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