Integral of $$$p^{6} \ln\left(p\right)$$$
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Find $$$\int p^{6} \ln\left(p\right)\, dp$$$.
Solution
For the integral $$$\int{p^{6} \ln{\left(p \right)} d p}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\ln{\left(p \right)}$$$ and $$$\operatorname{dv}=p^{6} dp$$$.
Then $$$\operatorname{du}=\left(\ln{\left(p \right)}\right)^{\prime }dp=\frac{dp}{p}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{p^{6} d p}=\frac{p^{7}}{7}$$$ (steps can be seen »).
The integral becomes
$${\color{red}{\int{p^{6} \ln{\left(p \right)} d p}}}={\color{red}{\left(\ln{\left(p \right)} \cdot \frac{p^{7}}{7}-\int{\frac{p^{7}}{7} \cdot \frac{1}{p} d p}\right)}}={\color{red}{\left(\frac{p^{7} \ln{\left(p \right)}}{7} - \int{\frac{p^{6}}{7} d p}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(p \right)}\, dp = c \int f{\left(p \right)}\, dp$$$ with $$$c=\frac{1}{7}$$$ and $$$f{\left(p \right)} = p^{6}$$$:
$$\frac{p^{7} \ln{\left(p \right)}}{7} - {\color{red}{\int{\frac{p^{6}}{7} d p}}} = \frac{p^{7} \ln{\left(p \right)}}{7} - {\color{red}{\left(\frac{\int{p^{6} d p}}{7}\right)}}$$
Apply the power rule $$$\int p^{n}\, dp = \frac{p^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=6$$$:
$$\frac{p^{7} \ln{\left(p \right)}}{7} - \frac{{\color{red}{\int{p^{6} d p}}}}{7}=\frac{p^{7} \ln{\left(p \right)}}{7} - \frac{{\color{red}{\frac{p^{1 + 6}}{1 + 6}}}}{7}=\frac{p^{7} \ln{\left(p \right)}}{7} - \frac{{\color{red}{\left(\frac{p^{7}}{7}\right)}}}{7}$$
Therefore,
$$\int{p^{6} \ln{\left(p \right)} d p} = \frac{p^{7} \ln{\left(p \right)}}{7} - \frac{p^{7}}{49}$$
Simplify:
$$\int{p^{6} \ln{\left(p \right)} d p} = \frac{p^{7} \left(7 \ln{\left(p \right)} - 1\right)}{49}$$
Add the constant of integration:
$$\int{p^{6} \ln{\left(p \right)} d p} = \frac{p^{7} \left(7 \ln{\left(p \right)} - 1\right)}{49}+C$$
Answer
$$$\int p^{6} \ln\left(p\right)\, dp = \frac{p^{7} \left(7 \ln\left(p\right) - 1\right)}{49} + C$$$A