The integral $\int x ^ { 3 } \ln x d x$ is

A) $\frac { x ^ { 4 } ( 4 \ln x + 1 ) } { 16 } + C$
B) $- \frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 16 } + C$
C) $- \frac { x ^ { 4 } ( 4 \ln x + 1 ) } { 16 } + C$
D) $\frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 8 } + C$
E) $\frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 16 } + C$

Question

The integral $\int x ^ { 3 } \ln x d x$ is

A) $\frac { x ^ { 4 } ( 4 \ln x + 1 ) } { 16 } + C$
B) $- \frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 16 } + C$
C) $- \frac { x ^ { 4 } ( 4 \ln x + 1 ) } { 16 } + C$
D) $\frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 8 } + C$
E) $\frac { x ^ { 4 } ( 4 \ln x - 1 ) } { 16 } + C$

Quizplus · Accepted Answer

To solve the integral $\int x^3 \ln x dx$, we use integration by parts, letting $u = \ln x$ and $dv = x^3 dx$. This gives $du = \frac{1}{x} dx$ and $v = \frac{x^4}{4}$. Applying the formula for integration by parts, $\int u dv = uv - \int v du$, we get $\frac{x^4}{4} \ln x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx = \frac{x^4}{4} \ln x - \frac{1}{4} \int x^3 dx = \frac{x^4}{4} \ln x - \frac{x^4}{16}$, which simplifies to $\frac{x^4(4\ln x - 1)}{16} + C$.

The Integral ∫x3ln⁡xdx\int x ^ { 3 } \ln x d x∫x3lnxdx

The Integral $\int x ^ { 3 } \ln x d x$