Question

Find \(\int 6x(x^2+6)dx\).

(a) \(\frac{3x^4}{2}+18x^2+C\)

(b) \(\frac{3x^4}{2}-18x+C\)

(c) \(\frac{3x^4}{2}-18x^2+C\)

(d) \(\frac{3x^4}{2}+x^2+C\)

I had been asked this question in final exam.

My query is from Methods of Integration-1 in portion Integrals of Mathematics – Class 12

Answer 1

Right choice is (a) \(\frac{3x^4}{2}+18x^2+C\)

To explain I would say: Let x^2=t

Differentiating w.r.t x, we get

2x dx=dt

\(\int 6x(x^2+6)dx=3\int (t+6) dt\)

3\(\int (t+6)dt=3\left (\frac{t^2}{2}+6t\right )=\frac{3t^2}{2}+18t\)

Replacing t with x^2

\(\int 6x(x^2+6)dx=\frac{3x^4}{2}+18x^2+C\)