The value of integral -dx2x-31-32x-15-3 iswhere C is integration constant

I=∫dx(2x 3)^1/3(2x+1)^5/3 =∫dx(2x 3)^2·(2x+12x 3)^5/3 Taking 2x+12x 3=z⇒ 8(2x 3)^2dx=dz Now, nbsp; I= 18∫ z^ 5/3dz =316z^ 2/3+C ∴ I=316(2x 32x+1)^2/3+C ...

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Standard XII

Mathematics

Definition of Limit

The value of ...

Question

The value of integral $\int \frac{d x}{(2 x - 3)^{1 / 3} (2 x + 1)^{5 / 3}}$ is
(where $C$ is integration constant)

\frac{3}{16} {(\frac{2 x + 1}{2 x - 3})}^{2 / 3} + C

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\frac{3}{16} {(\frac{2 x - 3}{2 x + 1})}^{2 / 3} + C

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\frac{3}{2} {(\frac{2 x - 3}{2 x + 1})}^{2 / 3} + C

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\frac{3}{2} {(\frac{2 x + 1}{2 x - 3})}^{2 / 3} + C

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Solution

The correct option is B $\frac{3}{16} {(\frac{2 x - 3}{2 x + 1})}^{2 / 3} + C$
$I = \int \frac{d x}{(2 x - 3)^{1 / 3} (2 x + 1)^{5 / 3}} = \int \frac{d x}{(2 x - 3)^{2} \cdot {(\frac{2 x + 1}{2 x - 3})}^{5 / 3}}$

Taking $\frac{2 x + 1}{2 x - 3} = z \Rightarrow \frac{- 8}{(2 x - 3)^{2}} d x = d z$
Now,
$I = - \frac{1}{8} \int z^{- 5 / 3} d z = \frac{3}{16} z^{- 2 / 3} + C$
$∴ I = \frac{3}{16} {(\frac{2 x - 3}{2 x + 1})}^{2 / 3} + C$

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The value of integral ∫dx(2x−3)1/3(2x+1)5/3 is (where C is integration constant)

The correct option is B 316(2x−32x+1)2/3+CI=∫dx(2x−3)1/3(2x+1)5/3 =∫dx(2x−3)2⋅(2x+12x−3)5/3 Taking 2x+12x−3=z⇒−8(2x−3)2dx=dz Now, I=−18∫z−5/3dz =316z−2/3+C ∴I=316(2x−32x+1)2/3+C

The value of integral $\int \frac{d x}{(2 x - 3)^{1 / 3} (2 x + 1)^{5 / 3}}$ is
(where $C$ is integration constant)