Integration of $x/\sqrt{x^2-7}$ using trigonometric substitution

Question

Didn't find this one here so I'm asking away:

I tried integrating by substitution by ended up with just $x + C$ which is clearly wrong.

My work, as requested:

Answer 1

• $$\dfrac{1}{2}\int\dfrac{1}{\sqrt{x^2-7}}(2xdx)=\dfrac{1}{2}\int\dfrac{1}{\sqrt{x^2-7}}d(x^2-7)=?$$

• $$\int\dfrac{x}{\sqrt{x^2-7}}dx=\dfrac{1}{7}\int\dfrac{x}{\sqrt{\left(\frac{x}{\sqrt{7}}\right)^2-1}}dx=\dfrac{1}{i}\int\dfrac{\frac{x}{\sqrt{7}}}{\sqrt{1-\left(\frac{x}{\sqrt{7}}\right)^2}}d\left(\frac{x}{\sqrt{7}}\right)\\ =\dfrac{1}{i}\left(\left(\frac{x}{\sqrt{7}}\right)\arcsin\left(\frac{x}{\sqrt{7}}\right)-\int \arcsin \left(\frac{x}{\sqrt{7}}\right)d\left(\frac{x}{\sqrt{7}}\right)\right)$$ Which is what you'd get if you wanted integration by trig substitution (I don't know why you'd want it, but that's not my business).

Answer 2

One way: Let $u=x^2-7$, then $du=2x\ dx.$

Trig Substitution way:
Let $x=\sqrt 7\sec\theta$. Then $dx=\sqrt 7\tan\theta\sec\theta.$ Also, $x^2=7\sec^2\theta.$ Plugging this in into the integral we have $$\sqrt 7\int\frac{\sqrt 7\sec\theta}{\sqrt{7(\sec^2\theta-1)}}(\tan\theta\sec\theta)d\theta$$ $$\sqrt 7\int\frac{\sec^2\theta\tan\theta}{\tan\theta}d\theta=\sqrt 7\int\sec^2\theta d\theta=\sqrt 7\tan\theta+C$$ Since $x=\sqrt 7\sec\theta$, then $\sec\theta=\frac{x}{\sqrt 7}=\frac{\text{hypotenuse}}{\text{adjacent}}$. This means that $\tan\theta=\frac{\sqrt{x^2-7}}{\sqrt 7}$. Thus the answer is $$\sqrt{x^2-7}+C$$

Answer 3

In this case, no need to use trigonometric substitution. In fact, $\int\dfrac{x}{\sqrt{x^2 - 7}}dx =\dfrac{1}{2}\int\dfrac{2x}{\sqrt{x^2 - 7}}dx = \dfrac{1}{2}\int\dfrac{d(x^2 - 7)}{\sqrt{x^2 - 7}}dx = \sqrt{x^2 - 7}$ + C

Remark $\int{\dfrac{1}{\sqrt{u}}}du = \int u^{-1/2}du = \dfrac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$.