{d x}{(x+a)^{9/7}(x-b)^{5/7}}=

Correct Answer is : 72(a+b)(x−bx+a)27+c7/2(a+b)≤ft(x-b/x+a)²/7}+c2(a+b)7(x+ax−b)72+c, where ccc is the constant of integration.

7/a+b≤ft(x-b/x+a)^{9/7}+c, where c is the constant of integration.

7/a+b≤ft(x-b/x+a)^{5/7}+c, where c is the constant of integration.

7/2(a+b)≤ft(x-b/x+a)²/7}+c, where c is the constant of integration.

7/a+b≤ft(x-b/x+a)¹/7}+c, where c is the constant of integration.

Correct Answer is : 72(a+b)(x−bx+a)27+c7/2(a+b)≤ft(x-b/x+a)²/7}+c2(a+b)7(x+ax−b)72+c, where ccc is the constant of integration.

Test Index

MHTCET 25 Apr 2025 Shift 2 Paper

Question : 130 of 150

Marks: +1, -0

\int \frac{d x}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}=

$\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^{\frac{9}{7}}+c$ , where c is the constant of integration.
$\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^{\frac{5}{7}}+c$ , where $c$ is the constant of integration.
$\frac{7}{2(a+b)}\left(\frac{x-b}{x+a}\right)^{\frac{2}{7}}+c$ , where $c$ is the constant of integration.
$\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^{\frac{1}{7}}+c$ , where $c$ is the constant of integration.

Solution:

Question:

Evaluate the integral:

\int \frac{dx}{(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}}}

Options:

\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^{\frac{9}{7}} + c

\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^{\frac{5}{7}} + c

\frac{7}{2(a+b)}\left(\frac{x-a}{x+b}\right)^{\frac{2}{7}} + c

\frac{7}{2(a+b)}\left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + c

(where

c

is the constant of integration)

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Solution:

Step 1: Use the substitution:

t = \frac{x-b}{x+a}

Differentiate with respect to

x

dt = \frac{(x+a) - (x-b)}{(x+a)^2} \, dx = \frac{a+b}{(x+a)^2} \, dx

So,

\frac{dx}{(x+a)^2} = \frac{dt}{a+b}

Step 2: Rewrite the integrand in terms of

t

Since

x-b = t(x+a)

(x+a)^{\frac{9}{7}}(x-b)^{\frac{5}{7}} = (x+a)^{\frac{9}{7}} \cdot t^{\frac{5}{7}} (x+a)^{\frac{5}{7}} = (x+a)^2 \cdot t^{\frac{5}{7}}

The integral becomes:

I = \int \frac{dx}{(x+a)^2 \cdot t^{\frac{5}{7}}}

Using the substitution:

I = \frac{1}{a+b} \int \frac{dt}{t^{\frac{5}{7}}} = \frac{1}{a+b} \int t^{-\frac{5}{7}} dt

Step 3: Integrate:

I = \frac{1}{a+b} \cdot \frac{t^{\frac{2}{7}}}{\frac{2}{7}} + c = \frac{7}{2(a+b)} \cdot t^{\frac{2}{7}} + c

Step 4: Substitute back

t = \frac{x-b}{x+a}

I = \frac{7}{2(a+b)} \left(\frac{x-b}{x+a}\right)^{\frac{2}{7}} + c

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Answer:

Option D.

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