How do you find the antiderivative of $int (arctan(sqrtx)) dx$ ?

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How do you find the antiderivative of $int (arctan(sqrtx)) dx$ ?

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Answer 1 · 2016-11-07T03:37:44

$(x+1)arctan(sqrtx)-sqrtx+C$

Explanation:

$I=intarctan(sqrtx)dx$

Integration by parts takes the form $intudv=uv-intvdu$ . So, for $intarctan(sqrtx)dx$ , we should let $u=arctan(sqrtx)$ and $dv=dx$ .

Differentiating $u$ gives $(du)/dx=1/(1+(sqrtx)^2)*d/dxsqrtx$ or $du=1/(2sqrtx(1+x))dx$ . From $dv=dx$ we integrate to show that $v=x$ . Thus:

$I=uv-intvdu$

$color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dx$

$color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dx$

Letting $t=sqrtx$ implies that $t^2=x$ and furthermore, that $2tdt=dx$ through differentiating. Thus:

$I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)$

$color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dt$

$color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dt$

$color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dt$

$color(white)I=xarctan(sqrtx)-intdt+arctan(t)$

$color(white)I=xarctan(sqrtx)-t+arctan(t)$

Since $t=sqrtx$ :

$I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)$

$color(white)I=(x+1)arctan(sqrtx)-sqrtx+C$

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How do you find the antiderivative of $int (arctan(sqrtx)) dx$ ?

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Explanation:

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