Mathematics

$$\displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$$ equals to


ANSWER

$$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$$


SOLUTION
$$I = \displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$$

$$\quad  = \displaystyle \int  \frac{2x}{2\sqrt{1 + x^{2}}}.\ln (x + \sqrt{1 + x^{2}}) dx$$ 

Let 
$$I_1 = \displaystyle \int  \frac{2x}{2\sqrt{1 + x^{2}}}.$$

Put $$1+x^2=t\implies2x dx=dt$$

$$I_1=\displaystyle \int \dfrac{1}{2}t^{\tfrac{-1}{2}}dt=t^{\tfrac{1}{2}}= \sqrt{1 + x^{2}}$$

Using integral by parts,
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$

$$I = \displaystyle \ln (x + \sqrt{1 + x^{2}}) \int \frac{2x}{2\sqrt{1 + x^{2}}} dx -\int \left[ \frac{d\{\ln (x + \sqrt{1 + x^{2}})\}}{dx}. \int \frac{2x}{2\sqrt{1 + x^{2}}} dx\right ] dx+c$$

$$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 +

x^{2}} -\int \left[ \frac{1}{x + \sqrt{1 + x^{2}}}. \left(1+\frac{x}{\sqrt{1+x^2}} \right).\sqrt{1 + x^{2}} \right ] dx+c$$

$$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 +

x^{2}} -\int dx+c=\ln (x + \sqrt{1 + x^{2}}). \sqrt{1 +

x^{2}} -x+c$$

$$\quad =\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$$
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Single Correct Hard Published on 17th 09, 2020
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