Question

Evaluate $\int x^3{\ln }^{2}xdx$
(where $C$ is constant of integration)

• A. $\frac{x^4}{4}({\ln }^{2}x-\frac{\ln x}{2}+\frac{1}{8})+C$
• B. $-\frac{x^4}{4}({\ln }^{2}x-\frac{\ln x}{2}+\frac{1}{8})+C$
• C. $\frac{x^4}{4}({\ln }^{2}x+\frac{\ln x}{2}+\frac{1}{8})+C$
• D. $-\frac{x^4}{4}({\ln }^{2}x-\frac{\ln x}{2}-\frac{1}{8})+C$

Answer 1

The correct option is A $\frac{x^4}{4}({\ln }^{2}x-\frac{\ln x}{2}+\frac{1}{8})+C$

Let $I=\int x^3{\ln }^{2}xdx$
Using ILATE, we get
$I=\left({\ln }^{2}x\right)\cdot \frac{x^4}{4}-\int \frac{x^4}{4}\cdot (2\cdot \ln x)\frac{1}{x}dx$ $\Rightarrow I=\frac{x^4{\ln }^{2}x}{4}-\frac{1}{2}\int x^3\ln xdx$
Again using ILATE, we get
$\Rightarrow I=\frac{x^4}{4}{\ln }^{2}x-\frac{1}{2}[\ln x\frac{x^4}{4}-\int \frac{x^4}{4}\cdot \frac{1}{x}dx]$ $\Rightarrow I=\frac{x^4}{4}{\ln }^{2}x-\frac{1}{8}\left(\ln x\right)x^4+\frac{1}{8}\cdot \frac{x^4}{4}+C$
$\Rightarrow I=\frac{x^4}{4}({\ln }^{2}x-\frac{\ln x}{2}+\frac{1}{8})+C$