The value of integral -exx-1-x2e2x-1dx iswhere C is constant of integration

Let I=∫e^x(x+1)√(x^2e^2x 1)dx substituting, xe^x=z ⇒ e^x(x+1)dx=dz So, integral reduced to, nbsp;I=∫dz√(z^2 1) =ln|z+√(z^2 1)|+C =ln|xe^x+√(x^2e^2x 1)|+C ...

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Standard XII

Mathematics

Combination

The value of ...

Question

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

ln | x e^{x} - \sqrt{x^{2} e^{2 x} - 1} | + C

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ln | x e^{x} + \sqrt{x^{2} e^{2 x} - 1} | + C

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ln | x e^{x} + \sqrt{x^{2} e^{2 x} - 1} | + C

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ln | x e^{2 x} + \sqrt{x^{2} e^{2 x} - 1} | + C

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Solution

The correct option is B $ln | x e^{x} + \sqrt{x^{2} e^{2 x} - 1} | + C$
Let $I = \int \frac{e^{x} (x + 1)}{\sqrt{x^{2} e^{2 x} - 1}} d x$
substituting, $x e^{x} = z$
$\Rightarrow e^{x} (x + 1) d x = d z$
So, integral reduced to, $I = \int \frac{d z}{\sqrt{z^{2} - 1}} = ln | z + \sqrt{z^{2} - 1} | + C = ln | x e^{x} + \sqrt{x^{2} e^{2 x} - 1} | + C$

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The value of integral ∫ex(x+1)√x2e2x−1dx is (where C is constant of integration)

The correct option is B ln|xex+√x2e2x−1|+CLet I=∫ex(x+1)√x2e2x−1dx substituting, xex=z ⇒ex(x+1)dx=dz So, integral reduced to, I=∫dz√z2−1 =ln|z+√z2−1|+C =ln|xex+√x2e2x−1|+C

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