Question

Evaluate: $\int \frac{{11}^{\frac{x}{2}}}{\sqrt{{11}^{-x}+{11}^x}}dx$

• A. ${\log }_{11}e.\log [{11}^x+\sqrt{1+{11}^{2x}}]+c$
• B. $\frac{1}{{\log }_{11}e}{\sin }^{-1}{11}^x+c$
• C. $\frac{{\cosh }^{-1}{11}^x}{\log 11}+c$
• D. ${\log }_{11}e\log |{11}^x+\sqrt{{11}^{2x}-1}|+c$

Answer 1

The correct option is A ${\log }_{11}e.\log [{11}^x+\sqrt{1+{11}^{2x}}]+c$

Let $I=\int \frac{{11}^{\frac{x}{2}}}{\sqrt{{11}^{-x}+{11}^x}}dx$

$I=\int \frac{{11}^x}{\sqrt{1+{11}^{2x}}}dx$

put ${11}^x=t$

so ${11}^x\log 11dx=dt$

$I=\frac{1}{\log 11}\int \frac{1}{\sqrt{1+t^2}}dt$

We know, $\int \frac{1}{\sqrt{a^2+x^2}}dt=\log |x+\sqrt{a^2+x^2}|+c$

$I={\log }_{11}e.\log |t+\sqrt{1+t^2}|+c$

Now, Put the value of $t$

$I={\log }_{11}e.\log |{11}^x+\sqrt{1+{11}^{2x}}|$