NIMCET 2014 — Mathematics PYQ

Concept

• Integration by parts: A method to find integrals of products. The formula is given by:

  $\int uvdx=u\int vdx-\int \left(\frac{du}{dx}\times \int vdx\right)dx+C$, where $u=u(x)$ and $v=v(x)$.

• ILATE rule: Usually, the preference order for choosing $u$ is based on Inverse, Logarithm, Algebraic, Trigonometric, and Exponent functions.

• Formula: $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$.

Calculation

Let $I=\int \frac{x{e}^x}{\sqrt{1+e^x}}dx$.

Take $1+e^x=t^2$ ... (1).

Differentiating both sides with respect to $x$, we get:

$e^{x}dx=2tdt$.

From equation (1), we have:

$e^x=t^2-1$

$\Rightarrow x=\log (t^2-1)$

Now, substituting these values into the integral:

$I=\int \frac{\log (t^2-1)}{\sqrt{t^2}}2tdt$

$=2\times \int \frac{\log (t^2-1)}{t}\times tdt$

$=2\int \log (t^2-1)dt$

Using the integration by parts rule:

$=2\left[\log (t^2-1)\times t-\int \frac{2t}{t^2-1}\times tdt\right]$

$=2t\log (t^2-1)-4\int \frac{t^2}{t^2-1}dt$

$=2t\log (t^2-1)-4\int \left[1+\frac{1}{t^2-1}\right]dt$

$=2t\log (t^2-1)-4t-4\times \frac{1}{2}\log \left(\frac{t-1}{t+1}\right)+C$

$=2t\log (t^2-1)-4t-2\log \left(\frac{t-1}{t+1}\right)+C$

$=2t(\log (t^2-1)-2)-2\log \left(\frac{t-1}{t+1}\right)+C$

Resubstitute the value of $t=\sqrt{1+e^x}$ and $\log (t^2-1)=x$:

$=2(x-2)\sqrt{1+e^x}-2\log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C$

$=(2x-4)\sqrt{1+e^x}-2\log \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C$

By comparing this result with the original equation, we find:

$f(x)=2x-4$

Correct Option: 2