Find ∫ 10 log⁡x.x^2 dx

Category: QuestionsFind ∫ 10 log⁡x.x^2 dx
Editor">Editor Staff asked 11 months ago

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
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This interesting question is from Integration by Parts topic in section Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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.