Question

Find ∫ 10 log⁡x.x^2 dx

(a) \(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)

(b) \(\frac{10x^3}{3} \left(log⁡x-\frac{x^3}{3}\right)+C\)

(c) \(-\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)

(d) \(\left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)

The question was asked in an interview for job.

This interesting question is from Integration by Parts topic in section Integrals of Mathematics – Class 12

Answer 1

Right answer is (a) \(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)

To explain I would say: ∫ 10 log⁡x.x^2 dx=10∫ log⁡x.x^2 dx

By using the formula, ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get

\(10\int log⁡x.x^2 \,dx=10(log⁡x \int x^2 \,dx-\int (log⁡x)’\int \,x^2 \,dx)\)

=\(10 \left(\frac{x^3 log⁡x}{3}-(\int \frac{1}{x}.\frac{x^3}{3} \,dx)\right)+C\)

=\(10 \left(\frac{x^3 log⁡x}{3}-\frac{1}{3} \frac{x^3}{3}\right)+C\)

=\(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\).