Evaluate the following integrals:

Assume x2 + 1 = t
⇒d(x2 + 1) = dt
⇒2x dx = dt
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x3 can be write as x2.x
∴ Now the given equation becomes
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x2 + 1 = t ⇒ x2 = t - 1
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But t = (x2 + 1)
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