Question

The value of $\int \frac{dx}{{(x+a)}^{\frac{8}{7}}{(x-b)}^{\frac{6}{7}}}$, is equal to

• A. $\frac{3}{2(a+b)}{\left(\frac{x+a}{x-b}\right)}^{\frac{2}{3}}+C$
• B. $\frac{3}{(a+b)}{\left(\frac{x-b}{x+a}\right)}^{\frac{1}{3}}+C$
• C. $\frac{7}{(a+b)}{\left(\frac{x-b}{x+a}\right)}^{\frac{1}{7}}+C$
• D. None of these

Answer 1

The correct option is C $\frac{7}{(a+b)}{\left(\frac{x-b}{x+a}\right)}^{\frac{1}{7}}+C$

$=\int \frac{1}{(x+a{)}^{1+\frac{1}{7}}(x-b{)}^{1-\frac{1}{7}}}$

$=\frac{(x+a)(x-b{)}^{\frac{1}{7}}}{(x-b)(x-a{)}^2(x+a{)}^{\frac{1}{7}}}$
Let,$\frac{(x-b)}{(x+a)}=t$

$(a+b)dx=(x+a{)}^{2}dt$
So,$=\int \frac{t^{\frac{-6}{7}}dt}{(a+b)}$

$=7.\frac{t^{\frac{1}{7}}}{(a+b)}+c$

$=\frac{7}{(a+b)}.(\frac{(x-b)}{(x+a)}{)}^{\frac{1}{7}}+c$