Find \int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx.

Category: QuestionsFind \int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx.
Editor">Editor Staff asked 11 months ago

Find \int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx.
(a) 4-\sqrt{2}
(b) 4+2\sqrt{2}
(c) 4-2\sqrt{2}
(d) 1-2\sqrt{2}
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My enquiry is from Evaluation of Definite Integrals by Substitution topic in portion Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

Correct choice is (c) 4-2\sqrt{2}
To explain I would say: I=\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx
Let x^5+3=t
Differentiating w.r.t x, we get
5x^4 dx=dt
The new limits
when x=-1,t=2
when x=1,t=4
∴\int_{-1}^1 \frac{5x^4}{\sqrt{x^5+3}} dx=\int_2^4 \frac{dt}{\sqrt{t}}