Question

Evaluate integral $\int e^{3a\log x}+e^{3x\log a}dx$

Answer 1

Solve the given integral

Given: $\int e^{3a\log x}+e^{3x\log a}dx$

We know that ${\left(a\right)}^{mn}={\left(a^m\right)}^n$

$$\begin{gathered} =\int \left[{\left(e^{\log x}\right)}^{3a}+{\left(e^{\log a}\right)}^{3x}\right]dx \\ =\int \left[x^{3a}+a^{3x}\right]dx\left[\because e^{\mathrm{logx}}=x\right] \\ =\frac{x^{3a+1}}{3a+1}+\frac{1}{3}\frac{a^{3x}}{\log a}+C\mathbf{}\left[\because \int x^n\mathrm{dx}=\frac{x^{n+1}}{n+1},\int a^{\mathrm{nx}}=\frac{a^x}{\mathrm{nloga}}\right] \end{gathered}$$

Hence the required answer is $\frac{{\mathbf{x}}^{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{1}}}{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{1}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{a}}^{\mathbf{3}\mathbf{x}}\mathbf{loga}$