Integral of $$$\ln\left(x \sqrt{x^{21}}\right)$$$
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Your Input
Find $$$\int \ln\left(x \sqrt{x^{21}}\right)\, dx$$$.
Solution
The input is rewritten: $$$\int{\ln{\left(x \sqrt{x^{21}} \right)} d x}=\int{\frac{23 \ln{\left(x \right)}}{2} d x}$$$.
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{23}{2}$$$ and $$$f{\left(x \right)} = \ln{\left(x \right)}$$$:
$${\color{red}{\int{\frac{23 \ln{\left(x \right)}}{2} d x}}} = {\color{red}{\left(\frac{23 \int{\ln{\left(x \right)} d x}}{2}\right)}}$$
For the integral $$$\int{\ln{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\ln{\left(x \right)}$$$ and $$$\operatorname{dv}=dx$$$.
Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d x}=x$$$ (steps can be seen »).
The integral can be rewritten as
$$\frac{23 {\color{red}{\int{\ln{\left(x \right)} d x}}}}{2}=\frac{23 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}}{2}=\frac{23 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}}{2}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\frac{23 x \ln{\left(x \right)}}{2} - \frac{23 {\color{red}{\int{1 d x}}}}{2} = \frac{23 x \ln{\left(x \right)}}{2} - \frac{23 {\color{red}{x}}}{2}$$
Therefore,
$$\int{\frac{23 \ln{\left(x \right)}}{2} d x} = \frac{23 x \ln{\left(x \right)}}{2} - \frac{23 x}{2}$$
Simplify:
$$\int{\frac{23 \ln{\left(x \right)}}{2} d x} = \frac{23 x \left(\ln{\left(x \right)} - 1\right)}{2}$$
Add the constant of integration:
$$\int{\frac{23 \ln{\left(x \right)}}{2} d x} = \frac{23 x \left(\ln{\left(x \right)} - 1\right)}{2}+C$$
Answer
$$$\int \ln\left(x \sqrt{x^{21}}\right)\, dx = \frac{23 x \left(\ln\left(x\right) - 1\right)}{2} + C$$$A