Question

Repeated application of integration by parts gives us the reduction formula, if the integrand is dependent on a natural number $n$.

If $\int \frac{dx}{x^n\sqrt{ax+b}}=-\frac{\sqrt{ax+b}}{(n-1)b{x}^{n-1}}-A\int \frac{dx}{x^{n-1}\sqrt{ax+b}}+C$, then $A$ is equal to

• A. $\frac{2n-3}{2n-1}$
• B. $\frac{2n-3}{2n-2}\frac{a}{b}$
• C. $\frac{n-2}{n-1}\frac{a}{b}$
• D. $\frac{n-3}{n-2}\frac{a}{b}$

Answer 1

The correct option is B $\frac{2n-3}{2n-2}\frac{a}{b}$

$\int \frac{dx}{x^n\left(\sqrt{ax+b}\right)}=-\frac{\sqrt{ax+b}}{(n-1)b{x}^{n-1}}-A\int \frac{dx}{x^{n-1}\left(\sqrt{ax+b}\right)}+C$

$\Rightarrow \int \frac{1+Ax}{x^n\left(\sqrt{ax+b}\right)}dx=-\frac{\sqrt{ax+b}}{(n-1)b{x}^{n-1}}+C$

Differentiating on both sides,

$\frac{1+Ax}{x^n\left(\sqrt{ax+b}\right)}=-\frac{x^{n-1}\frac{a}{2}(ax+b{)}^{-\frac{1}{2}}-(ax+b{)}^{\frac{1}{2}}(n-1)x^{n-2}}{b(n-1)x^{2n-2}}$

$\Rightarrow \frac{1+Ax}{x^n\sqrt{ax+b}}=-\frac{ax-2(n-1)(ax+b)}{b(n-1)2x^n\sqrt{ax+b}}$

Simplifying further, it gives

$A=\frac{2n-3}{2n-2}\frac{a}{b}$