Method 1.Let I = ∫2zdz/√(z^2+1) Let u=z^2+1, then du=2zdz I=∫ u^ 1/3du In the form of u^ndu = u^2/3/2/3+C Integrate with respect to u. = 3/2u^2/3+C=3/2( ...

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Standard XII

Physics

Antiderivative

Evaluate: ∫2z...

Question

Evaluate: $\int \frac{2 z d z}{\sqrt[3]{z^{2} + 1}}$

$\frac{3}{2} (z^{2} + 1)^{- 2 / 3} + c$

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$\frac{3}{2} (z^{2} + 1)^{2 / 3} + c$

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$\frac{3}{2} (z)^{4 / 3} + c$

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None of these

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Solution

The correct option is B
$\frac{3}{2} (z^{2} + 1)^{2 / 3} + c$

Method-1.Let I = $\int \frac{2 z d z}{\sqrt[3]{z^{2} + 1}}$

Let $u = z^{2} + 1$ , then $d u = 2 z d z$

$I = \int u^{- 1 / 3} d u$ In the form of $u^{n} d u$

= $\frac{u^{2 / 3}}{2 / 3} + C$ Integrate with respect to u.

= $\frac{3}{2} u^{2 / 3} + C = \frac{3}{2} (z^{2} + 1)^{2 / 3} + C$
( Replace u by $z^{2} + 1$ )

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Evaluate: ∫2zdz3√z2+1

The correct option is B 32(z2+1)2/3+c Method-1.Let I = ∫2zdz3√z2+1 Let u=z2+1, then du=2zdz I=∫u−1/3du In the form of undu = u2/32/3+C Integrate with respect to u. = 32u2/3+C=32(z2+1)2/3+C ( Replace u by z2+1)

Evaluate: $\int \frac{2 z d z}{\sqrt[3]{z^{2} + 1}}$