<s>[INST] Trigonometric Substitution (integration) $$\int\frac{dx}{(x^{2}-36)^{3/2}}$$
My attempt:
the factor in the denominator implies
$$x^{2}-36=x^{2}-6^{2}$$
substituting $x=6\sec\theta$, noting that $dx=6\tan\theta \sec\theta$ 
$$x^{2}-6^{2}=6^{2}\sec^{2}\theta-6^{2}=6^{2}\tan^{2}\theta$$
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\int\frac{6\tan\theta \sec\theta}{36\tan^{2}\theta}=\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}$$
using trig identities:
$$\frac{1}{6}\int\frac{\sec\theta}{\tan\theta}=\frac{1}{6}\int \sin^{-1}\theta$$
now using integration by parts:
$$\frac{1}{6}\int \sin^{-1}\theta$$
$$u=\sin^{-1}\theta, du=\frac{1}{\sqrt{1-\theta^{2}}}, dv=1, v=\theta$$
using $uv-\int{vdu}$
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\int{\frac{\theta}{\sqrt{1-\theta^{2}}}}d\theta\bigg)$$
now using simple substitution:$$z=1-\theta^{2}, dz=-2\theta d\theta, -\frac{1}{2}du=\theta d\theta$$
it is apparent that
$$\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}\int{\frac{dz}{\sqrt{z}}}\bigg)\bigg)$$
$$=\frac{1}{6}\bigg(\theta \sin^{-1}\theta-\bigg(-\frac{1}{2}(2\sqrt{z})\bigg)\bigg)=\frac{1}{6}\bigg(\theta \sin^{-1}\theta+\sqrt{1+\theta^{2}}\bigg)$$
$$=\frac{1}{6}\theta \sin^{-1}\theta+\frac{1}{6}\sqrt{1+\theta^{2}}+C$$
I have the following questions:
1.This integral seems tricky and drawn out to me, is there another method that reduces the steps/ methods of integration? I had to use trig substitution, integration by parts, and substitution in order to solve the integral, what can I do to find easier ways to complete integrals of this type?
2.Is this solution even correct? wolfram alpha says the solution to this integral is $-\frac{x}{36\sqrt{x^{2}-36}}+C$ how can i determine equivalence?
 [/INST] Without substitution:
$$\int\frac{dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{(x^2-(x^2-36))\,dx}{(x^{2}-36)^{3/2}}=\frac1{36}\int\frac{x^2\,dx}{(x^{2}-36)^{3/2}}-\frac1{36}\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
Then by parts on the first term,
$$\int\frac{x\cdot x\,dx}{(x^{2}-36)^{3/2}}=-\frac x{(x^{2}-36)^{1/2}}+\int\frac{dx}{(x^{2}-36)^{1/2}}.$$
After scaling the variable with a factor $6$, you recognize a known derivative,
$$\int\frac{dx}{\sqrt{x^2-1}}=\text{arcosh}(x).$$
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