evaluate integral $\int_3^{11} \frac{dx}{(x+5)\sqrt{1+x}} $

Question

How to evaluate
$$\int_3^{11} \frac{dx}{(x+5)\sqrt{1+x}} $$

It is a common device in such integrals to write $t^2$ for the expression under the square root.

Thus, $x = t^2-1$, so that $dx = 2t\ dt$

I got stuck on transitioning to:

$$\int_3^{11} \frac{dx}{(x+5)\sqrt{1+x}} = \int_2^{2\sqrt3} \frac{2tdt}{(t^2+4)t} $$

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(Above has been completed)

Then how do I transition to:

when, $$2 tan \theta$$ so that, $$2 sec^2 \theta d \theta$$

as t rises from $2$ to $2\sqrt2$, $\theta$ rises from $\frac{1}{4}\pi$ to $\frac{1}{3}\pi$

I want $$\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi}$$

However do not know how to get these boundaries. I have tried placed 2 in $t$ but this doesn't give me the answer.

Answer 1

Keep going:$$\int_3^{11} \frac{dx}{(x+5)\sqrt{1+x}}=\int_2^{2\sqrt3}\frac{2tdt}{(t^2+4)t}=\int_2^{2\sqrt3}\dfrac{2dt}{t^2+4} $$

Can you take it from here?

Answer 2

Let $x+1=t^2$ then $\sqrt{x+1}=\lvert t\rvert$, $dx = 2tdt$, $x+5=t^2+4$. We need to change limits of integral : if $x=3$ then $t=\sqrt{x+1}=\sqrt{3+1}=2$ and if $x=11$ then $t=\sqrt{4*3}=2\sqrt{3}$. And if $t$ changes from $2$ to $2\sqrt{3}$ then $\lvert t \rvert=t$.

Then after substitution you get the right side of your equivalence.

Answer 3

You may continue with

$$I= \int_2^{2\sqrt3} \frac{2tdt}{(t^2+4)t} =\frac12 \int_2^{2\sqrt3} \frac{dt}{(\frac t2)^2+1} $$

and make the substitution $\tan\theta =\frac t2$ along with $dt = 2\sec^2\theta d\theta$. Then, the lower limit is $\tan^{-1} 1 = \frac \pi4$ and the upper limit $\tan^{-1} \sqrt3= \frac \pi3$. As a result, the integral becomes,

$$I=\int_{\pi/4}^{\pi/3}\frac {\sec^2\theta}{\tan^2\theta+1}d\theta =\int_{\pi/4}^{\pi/3}d\theta=\frac\pi3-\frac\pi4=\frac\pi{12}$$