# 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}}

=[2\sqrt{t}]_2^4=2(\sqrt{4}-\sqrt{2})=4-2\sqrt{2}