Question

$\int ta{n}^{-1}\sqrt{x}dx$ is equal to

• A. $\left(x+1\right)ta{n}^{-1}\sqrt{x}-\sqrt{x}+C$
• B. $xta{n}^{-1}\sqrt{x}-\sqrt{x}+C$
• C. $\sqrt{x}-xta{n}^{-1}\sqrt{x}+C$
• D. $\sqrt{x}-(x+1)ta{n}^{-1}\sqrt{x}+C$

Answer 1

Answer: (A)

We have, $I=\int 1\cdot ta{n}^{-1}\sqrt{x}dx$

$\Rightarrow I=ta{n}^{-1}\sqrt{x}\cdot \left(x\right)-\int \frac{1}{1+x}\times \frac{1}{2\sqrt{x}}\times xdx$

$=xta{n}^{-1}\sqrt{x}-\int \frac{x}{\left(1+x\right)2\sqrt{x}}dx$

$=xta{n}^{-1}\sqrt{x}-\int \left(\frac{1+x}{\left(1+x\right)2\sqrt{x}}-\frac{1}{\left(1+x\right)2\sqrt{x}}\right)dx$

$=xta{n}^{-1}\sqrt{x}-\int \frac{dx}{2\sqrt{x}}+\int \frac{dx}{2\sqrt{x}\left(1+x\right)}$

$=xta{n}^{-1}\sqrt{x}-\sqrt{x}+ta{n}^{-1}\sqrt{x}+C$

$=\left(x+1\right)ta{n}^{-1}\sqrt{x}-\sqrt{x}+C$