Find the integral using integration by parts. $$\int x ^ { 4 } \ln x d x$$
A) $\int x ^ { 4 } \ln x d x = \frac { 1 } { 25 } ( 5 \ln x - x ) + C$
B) $\int x ^ { 4 } \ln x d x = x ^ { 5 } ( 5 \ln x - 1 ) + C$
C) $\int x ^ { 4 } \ln x d x = \frac { 1 } { 25 } x ^ { 5 } ( 5 \ln x - 1 ) + C$
D) $\int x ^ { 4 } \ln x d x = \frac { 1 } { 25 } x ^ { 5 } ( 5 \ln x - x ) + C$
E) $\int x ^ { 4 } \ln x d x = \frac { 1 } { 25 } ( \ln x - 1 ) + C$

Question

Quizplus · Accepted Answer

We use integration by parts with $u=\ln x$ and $dv=x^4dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{5}x^5$, so $$\int x^4\ln xdx=uv-\int vdu=\frac{1}{5}x^5\ln x-\int \frac{1}{5}x^4dx=\frac{1}{5}x^5\ln x-\frac{1}{25}x^5+C=\boxed{\frac{1}{25}x^5(5\ln x-1)+C}.$$

Find the Integral Using Integration by Parts ∫x4ln⁡xdx\int x ^ { 4 } \ln x d x∫x4lnxdx

Find the Integral Using Integration by Parts $\int x ^ { 4 } \ln x d x$