Use integration by parts to evaluate $\int 3 x ^ { 3 } \ln x d x$ .
A) $\frac { - 3 x ^ { 4 } ( 4 \ln ( x ) + 1 ) } { 4 } + C$
B) $\frac { 3 x ^ { 3 } ( 3 \ln ( x ) + 1 ) } { 3 } + C$
C) $\frac { 3 x ^ { 4 } ( 4 \ln ( x ) + 1 ) } { 16 } + C$
D) $\frac { 3 x ^ { 3 } ( 3 \ln ( x ) - 1 ) } { 9 } + C$
E) $\frac { 3 x ^ { 4 } ( 4 \ln ( x ) - 1 ) } { 16 } + C$

Question

Quizplus · Accepted Answer

Using integration by parts, let $ u = \ln x $ and $ dv = 3x^3 dx $. Then $ du = \frac{1}{x} dx $ and $ v = \frac{3}{4}x^4 $. Applying the integration by parts formula $\int u \, dv = uv - \int v \, du$, we get $\frac{3}{4}x^4 \ln x - \int \frac{3}{4}x^4 \cdot \frac{1}{x} dx = \frac{3}{4}x^4 \ln x - \frac{3}{16}x^4 + C$, which simplifies to $\frac{3x^4(4\ln x - 1)}{16} + C$.

Use Integration by Parts to Evaluate ∫3x3ln⁡xdx\int 3 x ^ { 3 } \ln x d x∫3x3lnxdx

Use Integration by Parts to Evaluate $\int 3 x ^ { 3 } \ln x d x$