Test Index

Integral Calculus

© examsnet.com

Question : 56 of 73

Marks: +1, -0

Integrate

x^{3} \log x

$\frac{x^{4}}{16}+k$
$\frac{x^{4}}{16}(4 \log x-1)+k$
$4 \log x-1+k$
None of these

Solution:

Concept:

Use integration by parts:

\int u \, dv = uv - \int v \, du

.

Explanation:

Let

u = \log x

and

dv = x^3 \, dx

.

Then

du = \frac{1}{x} \, dx

and

v = \frac{x^4}{4}

.

Apply integration by parts:

\int x^3 \log x \, dx = (\log x) \cdot \frac{x^4}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx

= \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx

= \frac{x^4}{4} \log x - \frac{1}{4} \cdot \frac{x^4}{4} + C

= \frac{x^4}{4} \log x - \frac{x^4}{16} + C

= \frac{x^4}{16} (4 \log x - 1) + C

.

Answer:

\frac{x^4}{16} (4 \log x - 1) + k

(Option B).

© examsnet.com

Go to Question:

Prev Question

Next Question