How do you evaluate $\int$ $int$ $\frac{arctan (\sqrt{x})}{\sqrt{x}}$ $arctan(sqrt(x))/sqrt(x)$ dx?

Question

Use the Table of Integrals to evaluate the integral. (Use C for the constant of integration.)

$\int$ $int$ $\frac{arctan (\sqrt{x})}{\sqrt{x}}$ $arctan(sqrt(x))/sqrt(x)$ dx

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Answer 1 · 2017-06-15T22:43:00

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Use the u substitution.

u = $\sqrt{x}$ $sqrt(x)$

du = $\frac{1}{2 \sqrt{x}}$ $1/(2sqrt(x))$ dx

2du = $\frac{1}{\sqrt{x}}$ $1/sqrt(x)$ dx

Write the new formula after the u substitution.

2 $\int$ $int$ ${tan}^{- 1} (u)$ $tan^-1(u)$ du

Use table 89 to find the integral of 2 ${tan}^{- 1} (u)$ $tan^-1(u)$ .

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2 $\int$ $int$ ${tan}^{- 1} (u)$ $tan^-1(u)$ du
= 2[u ${tan}^{- 1} (u)$ $tan^-1(u)$ - $\frac{1}{2}$ $1/2$ ln(1 + $u^{2}$ $u^2$ )] + C

Replace the u variable back in the terms of x.

= 2[ $\sqrt{x}$ $sqrt(x)$ ${tan}^{- 1} (\sqrt{x})$ $tan^-1(sqrt(x))$ - $\frac{1}{2}$ $1/2$ ln(1 + ${\sqrt{x}}^{2}$ $sqrt(x)^2$ )] + C

Simplify the answer.

= 2[ $\sqrt{x}$ $sqrt(x)$ ${tan}^{- 1} (\sqrt{x})$ $tan^-1(sqrt(x))$ - $\frac{1}{2}$ $1/2$ ln(1 + $x$ $x$ )] + C

= 2 $\sqrt{x}$ $sqrt(x)$ ${tan}^{- 1} (\sqrt{x})$ $tan^-1(sqrt(x))$ - ln(1 + $x$ $x$ ) + C