How do you find the antiderivative of $\int (arctan (\sqrt{x})) d x$ $int (arctan(sqrtx)) dx$ ?

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How do you find the antiderivative of $\int (arctan (\sqrt{x})) d x$ $int (arctan(sqrtx)) dx$ ?

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Answer 1 · 2016-11-07T03:37:44

$(x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $(x+1)arctan(sqrtx)-sqrtx+C$

Explanation:

$I = \int arctan (\sqrt{x}) d x$ $I=intarctan(sqrtx)dx$

Integration by parts takes the form $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ . So, for $\int arctan (\sqrt{x}) d x$ $intarctan(sqrtx)dx$ , we should let $u = arctan (\sqrt{x})$ $u=arctan(sqrtx)$ and $d v = d x$ $dv=dx$ .

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

$I = u v - \int v d u$ $I=uv-intvdu$

$I = x arctan (\sqrt{x}) - \int \frac{x}{2 \sqrt{x} (1 + x)} d x$ $color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dx$

$I = x arctan (\sqrt{x}) - \frac{1}{2} \int \frac{\sqrt{x}}{1 + x} d x$ $color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dx$

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

$I = x arctan (\sqrt{x}) - \frac{1}{2} \int \frac{t}{1 + t^{2}} (2 t d t)$ $I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)$

$I = x arctan (\sqrt{x}) - \int \frac{t^{2}}{1 + t^{2}} d t$ $color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dt$

$I = x arctan (\sqrt{x}) - \int \frac{1 + t^{2} - 1}{1 + t^{2}} d t$ $color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dt$

$I = x arctan (\sqrt{x}) - \int \frac{1 + t^{2}}{1 + t^{2}} d t + \int \frac{1}{1 + t^{2}} d t$ $color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dt$

$I = x arctan (\sqrt{x}) - \int d t + arctan (t)$ $color(white)I=xarctan(sqrtx)-intdt+arctan(t)$

$I = x arctan (\sqrt{x}) - t + arctan (t)$ $color(white)I=xarctan(sqrtx)-t+arctan(t)$

Since $t = \sqrt{x}$ $t=sqrtx$ :

$I = x arctan (\sqrt{x}) - \sqrt{x} + arctan (\sqrt{x})$ $I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)$

$I = (x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $color(white)I=(x+1)arctan(sqrtx)-sqrtx+C$

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How do you find the antiderivative of $\int (arctan (\sqrt{x})) d x$ $int (arctan(sqrtx)) dx$ ?

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How do you find the antiderivative of int (arctan(sqrtx)) dx∫(arctan(√x))dxint (arctan(sqrtx)) dx?

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How do you find the antiderivative of $\int (arctan (\sqrt{x})) d x$ $int (arctan(sqrtx)) dx$ ?