Integral of $$$22 i a^{2} b^{x - 1} n t x$$$ with respect to $$$x$$$

Find $$$\int 22 i a^{2} b^{x - 1} n t x\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=22 i a^{2} n t$$$ and $$$f{\left(x \right)} = b^{x - 1} x$$$:

$${\color{red}{\int{22 i a^{2} b^{x - 1} n t x d x}}} = {\color{red}{\left(22 i a^{2} n t \int{b^{x - 1} x d x}\right)}}$$

For the integral $$$\int{b^{x - 1} x d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=b^{x - 1} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{b^{x - 1} d x}=\frac{b^{x - 1}}{\ln{\left(b \right)}}$$$ (steps can be seen »).

So,

$$22 i a^{2} n t {\color{red}{\int{b^{x - 1} x d x}}}=22 i a^{2} n t {\color{red}{\left(x \cdot \frac{b^{x - 1}}{\ln{\left(b \right)}}-\int{\frac{b^{x - 1}}{\ln{\left(b \right)}} \cdot 1 d x}\right)}}=22 i a^{2} n t {\color{red}{\left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \int{\frac{b^{x - 1}}{\ln{\left(b \right)}} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{\ln{\left(b \right)}}$$$ and $$$f{\left(x \right)} = b^{x - 1}$$$:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - {\color{red}{\int{\frac{b^{x - 1}}{\ln{\left(b \right)}} d x}}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - {\color{red}{\frac{\int{b^{x - 1} d x}}{\ln{\left(b \right)}}}}\right)$$

Let $$$u=x - 1$$$.

Then $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{x - 1} d x}}}}{\ln{\left(b \right)}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{u} d u}}}}{\ln{\left(b \right)}}\right)$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=b$$$:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\int{b^{u} d u}}}}{\ln{\left(b \right)}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{{\color{red}{\frac{b^{u}}{\ln{\left(b \right)}}}}}{\ln{\left(b \right)}}\right)$$

Recall that $$$u=x - 1$$$:

$$22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{{\color{red}{u}}}}{\ln{\left(b \right)}^{2}}\right) = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{{\color{red}{\left(x - 1\right)}}}}{\ln{\left(b \right)}^{2}}\right)$$

Therefore,

$$\int{22 i a^{2} b^{x - 1} n t x d x} = 22 i a^{2} n t \left(\frac{b^{x - 1} x}{\ln{\left(b \right)}} - \frac{b^{x - 1}}{\ln{\left(b \right)}^{2}}\right)$$

Simplify:

$$\int{22 i a^{2} b^{x - 1} n t x d x} = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln{\left(b \right)} - 1\right)}{\ln{\left(b \right)}^{2}}$$

Add the constant of integration:

$$\int{22 i a^{2} b^{x - 1} n t x d x} = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln{\left(b \right)} - 1\right)}{\ln{\left(b \right)}^{2}}+C$$

$$$\int 22 i a^{2} b^{x - 1} n t x\, dx = \frac{22 i a^{2} b^{x - 1} n t \left(x \ln\left(b\right) - 1\right)}{\ln^{2}\left(b\right)} + C$$$A