Let a, b, and c be positive constants. The value of ‘a’ in terms of ‘c’ if the value of integral ${\int }_0^1 \left({\mathrm{acx}}^{\mathrm{b}+1}+{\mathrm{a}}^3{\mathrm{bx}}^{3\mathrm{b}+5}\right)\mathrm{dx}$ is independent of b, equals:

$\begin{array}{l}\mathrm{I}={\int }_0^1 \left({\mathrm{acx}}^{\mathrm{b}+1}+{\mathrm{a}}^3{\mathrm{bx}}^{3\mathrm{b}+5}\right)\mathrm{dx}\\ ={\left[\mathrm{ac}\cdot \frac{{\mathrm{x}}^{\mathrm{b}+2}}{\mathrm{b}+2}+\frac{{\mathrm{a}}^3{\mathrm{bx}}^{3\mathrm{b}+6}}{3\mathrm{b}+6}\right]}_0^1=\frac{\mathrm{ac}}{\mathrm{b}+2}+\frac{{\mathrm{a}}^3\mathrm{b}}{3(\mathrm{b}+2)}\\ \mathrm{I}\left(\mathrm{b}\right)=\frac{1}{3(\mathrm{b}+2)}\left[3\mathrm{ac}+{\mathrm{a}}^3\mathrm{b}\right]\end{array}$

Hence, 

$\Rightarrow \frac{{\mathrm{a}}^3\left(\mathrm{b}+\frac{3\mathrm{c}}{{\mathrm{a}}^2}\right)}{3(\mathrm{b}+2)}$

If this is dependent of b, then $\frac{3\mathrm{c}}{{\mathrm{a}}^2}=2$

Alternatively: if the integral is independent of b, then $\mathrm{I}\text{'}\left(\mathrm{b}\right)=0$

$\begin{array}{l}\Rightarrow \mathrm{I}\text{'}\left(\mathrm{b}\right)=\frac{-\mathrm{ac}}{(\mathrm{b}+2{)}^2}+\mathrm{D}\left(\frac{{\mathrm{a}}^3}{3}\cdot \left(\frac{\mathrm{b}+2-2}{\mathrm{b}+2}\right)\right)\\ =\frac{-\mathrm{ac}}{(\mathrm{b}+2{)}^2}+\frac{{\mathrm{a}}^3}{3}\left(\frac{2}{(\mathrm{b}+2{)}^2}\right)\\ \mathrm{\alpha }=\sqrt{\frac{3\mathrm{c}}{2}}\end{array}$