MHTCET 17 Apr 2026 Shift 1 Paper

Question : 131 of 150

Marks: +1, -0

\int x^{3} \log x d x=

Solution:

Concept:

Use integration by parts to integrate the product of a polynomial and a logarithmic function.

Explanation:

Let

u = \log x

and

dv = x^3 \, dx

Then

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

and

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

Applying integration by parts:

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

Therefore:

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

Simplify the integral:

= \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

Factor out

\frac{x^4}{16}

= \frac{x^4}{16} [4 \log x - 1] + c

Answer:

Option A:

\frac{x^4}{16}[4 \log x - 1] + c

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