How do you integrate int(dx/(sqrt(2x^3 + 2x + 5)))(dx2x3+2x+5)?

1 Answer
Apr 3, 2015

The integral you wrote is impossible to do using "elementary functions" (at least according to the computer program Mathematica...it requires "Elliptic Functions").

I think you probably meant to do \int(\frac{dx}{\sqrt{2x^{2}+2x+5}})(dx2x2+2x+5). The first steps in doing this integral is to use a bit of tricky algebra (including the technique of completing the square) to get:

\int\frac{dx}{\sqrt{2x^{2}+2x+5}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{5}{2}}}dx2x2+2x+5=12dxx2+x+52

=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{x^{2}+x+\frac{1}{4}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}=12dxx2+x+14+94=12dx(x+12)2+94.

Now do the trigonometric substitution x+\frac{1}{2}=\frac{3}{2}\tan(\theta)x+12=32tan(θ) do so that \theta=\arctan(\frac{2}{3}x+\frac{1}{3})θ=arctan(23x+13) and dx=\frac{3}{2}\sec^{2}(\theta)d\thetadx=32sec2(θ)dθ.

Then the integral becomes:

\frac{1}{\sqrt{2}}\int\frac{dx}{\sqrt{(x+\frac{1}{2})^{2}+\frac{9}{4}}}=\frac{1}{\sqrt{2}}\int\frac{\frac{3}{2}\sec^{2}(\theta)d\theta}{\sqrt{\frac{9}{4}(\tan^{2}(\theta)+1)}}12dx(x+12)2+94=1232sec2(θ)dθ94(tan2(θ)+1), which simplifies to

\frac{1}{\sqrt{2}}\int\frac{\sec^{2}(\theta)d\theta}{\sqrt{\sec^{2}(\theta)}}12sec2(θ)dθsec2(θ)

after using a trigonometric identity. Technically-speaking, \sqrt{\sec^{2}(\theta)}=|sec(\theta)|sec2(θ)=|sec(θ)|, however, it is fine to get rid of the absolute value signs for the purpose of finishing the integral (assume \sec(\theta)\geq 0sec(θ)0).

Hence, the integral becomes \frac{1}{\sqrt{2}}\int\sec(\theta)d\theta12sec(θ)dθ. This last integral looks simple, but it is still tricky to do. One trick that works is to multiply it by \frac{\sec(\theta)+\tan(\theta}}{\sec(\theta)+\tan(\theta)sec(θ)+tan(θ}sec(θ)+tan(θ) and then do a substitution (you can try that if you want). Wolfram Alpha gives the answer as \frac{1}{\sqrt{2}}(ln(cos(\theta/2)+sin(\theta/2))-\ln(\cos(\theta/2)-\sin(\theta/2)))+C12(ln(cos(θ2)+sin(θ2))ln(cos(θ2)sin(θ2)))+C. Now just replace \thetaθ by \theta=\arctan(\frac{2}{3}x+\frac{1}{3})θ=arctan(23x+13) and you are done:

\int(\frac{dx}{\sqrt{2x^{2}+2x+5}})=\frac{1}{\sqrt{2}}(ln(cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))+sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3})))-\ln(\cos(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))-\sin(\frac{1}{2}\arctan(\frac{2}{3}x+\frac{1}{3}))))+C

It is possible to simplify this by using half angle formulas and the following equations, but it may not be worth the trouble: \cos(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{1}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{3}{\sqrt{4x^{2}+4x+10}}

\sin(\arctan(\frac{2}{3}x+\frac{1}{3}))=\frac{\frac{2}{3}x+\frac{1}{3}}{\sqrt{1+(\frac{2}{3}x+\frac{1}{3})^{2}}}=\frac{2x+1}{\sqrt{4x^{2}+4x+10}}