Integral of $$$e^{\frac{\ln\left(a\right)}{\ln\left(b\right)}}$$$ with respect to $$$a$$$

Find $$$\int e^{\frac{\ln\left(a\right)}{\ln\left(b\right)}}\, da$$$.

Let $$$u=\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}$$$.

Then $$$du=\left(\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}\right)^{\prime }da = \frac{1}{a \ln{\left(b \right)}} da$$$ (steps can be seen »), and we have that $$$\frac{da}{a} = \ln{\left(b \right)} du$$$.

So,

$${\color{red}{\int{e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} d a}}} = {\color{red}{\int{e^{u} e^{u \ln{\left(b \right)}} \ln{\left(b \right)} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\ln{\left(b \right)}$$$ and $$$f{\left(u \right)} = e^{u} e^{u \ln{\left(b \right)}}$$$:

$${\color{red}{\int{e^{u} e^{u \ln{\left(b \right)}} \ln{\left(b \right)} d u}}} = {\color{red}{\ln{\left(b \right)} \int{e^{u} e^{u \ln{\left(b \right)}} d u}}}$$

Let $$$v=u \ln{\left(b \right)}$$$.

Then $$$dv=\left(u \ln{\left(b \right)}\right)^{\prime }du = \ln{\left(b \right)} du$$$ (steps can be seen »), and we have that $$$du = \frac{dv}{\ln{\left(b \right)}}$$$.

So,

$$\ln{\left(b \right)} {\color{red}{\int{e^{u} e^{u \ln{\left(b \right)}} d u}}} = \ln{\left(b \right)} {\color{red}{\int{\frac{e^{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}}}{\ln{\left(b \right)}} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{\ln{\left(b \right)}}$$$ and $$$f{\left(v \right)} = e^{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}}$$$:

$$\ln{\left(b \right)} {\color{red}{\int{\frac{e^{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}}}{\ln{\left(b \right)}} d v}}} = \ln{\left(b \right)} {\color{red}{\frac{\int{e^{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}} d v}}{\ln{\left(b \right)}}}}$$

Let $$$w=\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}$$$.

Then $$$dw=\left(\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}\right)^{\prime }dv = \frac{\ln{\left(b \right)} + 1}{\ln{\left(b \right)}} dv$$$ (steps can be seen »), and we have that $$$dv = \frac{\ln{\left(b \right)} dw}{\ln{\left(b \right)} + 1}$$$.

Therefore,

$${\color{red}{\int{e^{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}} d v}}} = {\color{red}{\int{\frac{e^{w} \ln{\left(b \right)}}{\ln{\left(b \right)} + 1} d w}}}$$

Apply the constant multiple rule $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ with $$$c=\frac{\ln{\left(b \right)}}{\ln{\left(b \right)} + 1}$$$ and $$$f{\left(w \right)} = e^{w}$$$:

$${\color{red}{\int{\frac{e^{w} \ln{\left(b \right)}}{\ln{\left(b \right)} + 1} d w}}} = {\color{red}{\frac{\ln{\left(b \right)} \int{e^{w} d w}}{\ln{\left(b \right)} + 1}}}$$

The integral of the exponential function is $$$\int{e^{w} d w} = e^{w}$$$:

$$\frac{\ln{\left(b \right)} {\color{red}{\int{e^{w} d w}}}}{\ln{\left(b \right)} + 1} = \frac{\ln{\left(b \right)} {\color{red}{e^{w}}}}{\ln{\left(b \right)} + 1}$$

Recall that $$$w=\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}$$$:

$$\frac{\ln{\left(b \right)} e^{{\color{red}{w}}}}{\ln{\left(b \right)} + 1} = \frac{\ln{\left(b \right)} e^{{\color{red}{\frac{v \left(\ln{\left(b \right)} + 1\right)}{\ln{\left(b \right)}}}}}}{\ln{\left(b \right)} + 1}$$

Recall that $$$v=u \ln{\left(b \right)}$$$:

$$\frac{\ln{\left(b \right)} e^{\frac{\left(\ln{\left(b \right)} + 1\right) {\color{red}{v}}}{\ln{\left(b \right)}}}}{\ln{\left(b \right)} + 1} = \frac{\ln{\left(b \right)} e^{\frac{\left(\ln{\left(b \right)} + 1\right) {\color{red}{u \ln{\left(b \right)}}}}{\ln{\left(b \right)}}}}{\ln{\left(b \right)} + 1}$$

Recall that $$$u=\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}$$$:

$$\frac{\ln{\left(b \right)} e^{\left(\ln{\left(b \right)} + 1\right) {\color{red}{u}}}}{\ln{\left(b \right)} + 1} = \frac{\ln{\left(b \right)} e^{\left(\ln{\left(b \right)} + 1\right) {\color{red}{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}}}}}{\ln{\left(b \right)} + 1}$$

Therefore,

$$\int{e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} d a} = \frac{e^{\frac{\left(\ln{\left(b \right)} + 1\right) \ln{\left(a \right)}}{\ln{\left(b \right)}}} \ln{\left(b \right)}}{\ln{\left(b \right)} + 1}$$

Simplify:

$$\int{e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} d a} = \frac{a e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} \ln{\left(b \right)}}{\ln{\left(b \right)} + 1}$$

Add the constant of integration:

$$\int{e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} d a} = \frac{a e^{\frac{\ln{\left(a \right)}}{\ln{\left(b \right)}}} \ln{\left(b \right)}}{\ln{\left(b \right)} + 1}+C$$

$$$\int e^{\frac{\ln\left(a\right)}{\ln\left(b\right)}}\, da = \frac{a e^{\frac{\ln\left(a\right)}{\ln\left(b\right)}} \ln\left(b\right)}{\ln\left(b\right) + 1} + C$$$A