How do you find the integral of $\int \frac{3}{2 \sqrt{x} (1 + x)}$ $int 3/(2sqrtx(1+x)$ ?

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How do you find the integral of $\int \frac{3}{2 \sqrt{x} (1 + x)}$ $int 3/(2sqrtx(1+x)$ ?

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Answer 1 · 2016-11-08T20:05:17

$3 arctan (\sqrt{x}) + C$ $3arctan(sqrtx)+C$

Explanation:

$I = \int \frac{3}{2 \sqrt{x} (1 + x)} d x = \frac{3}{2} \int \frac{d x}{\sqrt{x} (1 + x)}$ $I=int3/(2sqrtx(1+x))dx=3/2intdx/(sqrtx(1+x))$

If the substitution presented by Eric Sia is not immediately apparent, another substitution we can try is $u = \sqrt{x}$ $u=sqrtx$ . This implies that $d u = \frac{1}{2 \sqrt{x}} d x$ $du=1/(2sqrtx)dx$ which we already have in the integrand.

$I = 3 \int \frac{1}{1 + x} \frac{1}{2 \sqrt{x}} d x = 3 \int \frac{1}{1 + u^{2}} d u$ $I=3int1/(1+x)1/(2sqrtx)dx=3int1/(1+u^2)du$

At this point, you may recognize that this is the arctangent integral. However, since this is in the "Integration by Trigonometric Substitution" section, we can derive the arctangent integral here using a trig substitution. Let $u = tan θ$ $u=tantheta$ . Thus $d u = {sec}^{2} θ d θ$ $du=sec^2thetad theta$ .

$I = 3 \int \frac{1}{1 + {tan}^{2} θ} ({sec}^{2} θ d θ) = 3 \int d θ = 3 θ + C$ $I=3int1/(1+tan^2theta)(sec^2thetad theta)=3intd theta=3theta+C$

Note that we used $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ . Also, reverse the substitution of $u = tan θ$ $u=tantheta$ to show that $θ = arctan (u)$ $theta=arctan(u)$ . Since $u = \sqrt{x}$ $u=sqrtx$ , we see that $θ = arctan (\sqrt{x})$ $theta=arctan(sqrtx)$ .

$I = 3 arctan (\sqrt{x}) + C$ $I=3arctan(sqrtx)+C$

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How do you find the integral of $\int \frac{3}{2 \sqrt{x} (1 + x)}$ $int 3/(2sqrtx(1+x)$ ?

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How do you find the integral of $\int \frac{3}{2 \sqrt{x} (1 + x)}$ $int 3/(2sqrtx(1+x)$ ?