How do you evaluate $int$ $arctan(sqrt(x))/sqrt(x)$ dx?

Question

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

$int$ $arctan(sqrt(x))/sqrt(x)$ dx

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Answer 1 · 2017-06-15T22:43:00

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Use the u substitution.

u = $sqrt(x)$

du = $1/(2sqrt(x))$ dx

2du = $1/sqrt(x)$ dx

Write the new formula after the u substitution.

2 $int$ $tan^-1(u)$ du

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

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

Replace the u variable back in the terms of x.

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

Simplify the answer.

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

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