MHB 205.8.4.30. Int 24/(144x^2+1)^2 dx

[karush]

206.8.4.30. Int 24/(144x^2+1)^2 dx

206.8.4.30
$\displaystyle
I_30=\int \frac{24}{(144x^2+1)^2}=
\arctan\left(12x\right)+\dfrac{12x}{144x^2+1}+C$

So $x=12\tan\left({u}\right) \therefore du=12\sec^2 (u)du$

By the answer assume a trig subst.
Didn't want to try reduction formula:
Continue or is there better?

 

[MarkFL]

I would first write the integral as:

$$I_{30}=2\int\frac{12}{((12x)^2+1)^2}\,dx$$

Now, let:

$$12x=\tan(\theta)\implies 12\,dx=\sec^2(\theta)\,d\theta$$

and apply the Pythagorean identity:

$$\tan^2(\alpha)+1=\sec^2(\alpha)$$

and you now have:

$$I_{30}=2\int\frac{\sec^2(\theta)}{\sec^4(\theta)}\,d\theta=2\int\cos^2(\theta)\,d\theta=\int \cos(2\theta)+1\,d\theta$$

Now all that's left is to complete the integration and back-substitute for $\theta$. :)

 

[karush]

205.8.4.30

After fixing typos and plagiarizing..

$$\displaystyle
I_{30}=\int \frac{24}{(144x^2+1)^2} \, dx =
\arctan\left(12x\right)+\dfrac{12x}{144x^2+1}+C$$
rewrte
$$\displaystyle
I_{30}=2\int \frac{12}{((12x)^2+1)^2} \, dx $$
trig subst
$$12x=\tan\left({\theta}\right)
\therefore 12dx=\sec^2 (\theta)d\theta$$
$$\displaystyle I_{30}=2\int\frac{\sec^2(\theta)}{\sec^4(\theta)}\,d\theta=2\int\cos^2(\theta)\,d\theta=\int \cos(2\theta)+1\,d\theta$$

back subst $\theta=\arctan{(12x)}$

How do you back subst this ??
Isn't it $\cos\left({2\theta}\right)-1$

 

[MarkFL]

A double-angle identity for cosine is:

$$\cos(2\alpha)=2\cos^2(\alpha)-1$$

Now, when you integrate, you will get:

$$I_{30}=\frac{1}{2}\sin(2\theta)+\theta+C$$

Apply the double-angle identity for sine, and various other identities:

$$I_{30}=\sin(\theta)\cos(\theta)+\theta+C=\tan(\theta)\cos^2(\theta)+\theta+C=\frac{\tan(\theta)}{\sec^2(\theta)}+\theta+C=\frac{\tan(\theta)}{\tan^2(\theta)+1}+\theta+C$$

Now, back-substitute for $\theta$:

$$I_{30}=\frac{12x}{(12x)^2+1}+\arctan(12x)+C$$

 

[karush]

206.8.4.30
$$\displaystyle
I_{30}=\int \frac{24}{(144x^2+1)^2} \, dx =
\arctan\left(12x\right)+\dfrac{12x}{144x^2+1}+C$$
rewrte
$$\displaystyle
I_{30}=2\int \frac{12}{((12x)^2+1)^2} \, dx $$
trig subst
$$12x=\tan\left({\theta}\right)
\therefore 12dx=\sec^2 (\theta)d\theta$$
$$\displaystyle I_{30}=2\int\frac{\sec^2(\theta)}{\sec^4(\theta)}\,d\theta=2\int\cos^2(\theta)\,d\theta=\int \cos(2\theta)+1\,d\theta$$
$$\displaystyle I_{30}
=\sin(\theta)\cos(\theta)+\theta+C
=\tan(\theta)\cos^2(\theta)+\theta+C \\
=\frac{\tan(\theta)}{\sec^2(\theta)}+\theta+C
=\frac{\tan(\theta)}{\tan^2(\theta)+1}+\theta+C$$
back subst $\theta=\arctan{(12x)}$
$$\displaystyle I_{30}=\frac{12x}{(12x)^2+1}+\arctan(12x)+C$$