Integral of x / (x^2 + 3/4 )

[suzanka]

I'm trying to solve this integral and I can't find what error I'm making.

\(\displaystyle \displaystyle I = \int \frac{x}{(x^2+3/4)} {dx}\)

I substitute \(\displaystyle \displaystyle x = \frac{\sqrt3}{2}tan(t) \), so \(\displaystyle \displaystyle dt = \frac{\sqrt3}{2}\frac{1}{(cos(x))^2} \)


so I get \(\displaystyle \displaystyle I = \frac{3}{4}\int \frac{\tan(t)(\cos(t))^2}{\frac{3}{4}tan(t))^2 + \frac{3}{4}} {dt} = \) \(\displaystyle \displaystyle \int \frac{tan(t)}{cos(t)^2 (tan(t)^2 + 1)} {dt}\)

As \(\displaystyle \displaystyle tan(t)^2 + 1 = \frac{1}{cos(t)^2} \) the integral becomes

\(\displaystyle \displaystyle
I = \int tan(t) {dt} = \int \frac{sin(t)}{cos(t)} {dt} = -\int \frac{dcos(t)}{cos(t)} = - ln|(cos(t)| + C
\)

Substitution of t gives \(\displaystyle \displaystyle I = - ln |cos(\arctan(\frac{2x}{\sqrt3}))| = - ln | \frac{1}{ \sqrt( 1 + \frac{4x^2}{3 })} | \)

But my textbook says it's \(\displaystyle \displaystyle

\frac{ln( x^2+3/4) }{2}
\)


Thank you!

 

[daon2]

I'm trying to solve this integral and I can't find what error I'm making.


\(\displaystyle \displaystyle I = \int \frac{x}{(x^2+3/4)} {dx}\)


I substitute \(\displaystyle \displaystyle x = \frac{\sqrt3}{2}tan(t) \), so \(\displaystyle \displaystyle dt = \frac{\sqrt3}{2}\frac{1}{(cos(x))^2} \)




so I get \(\displaystyle \displaystyle I = \frac{3}{4}\int \frac{\tan(t)(\cos(t))^2}{\frac{3}{4}tan(t))^2 + \frac{3}{4}} {dt} = \) \(\displaystyle \displaystyle \int \frac{tan(t)}{cos(t)^2 (tan(t)^2 + 1)} {dt}\)


As \(\displaystyle \displaystyle tan(t)^2 + 1 = \frac{1}{cos(t)^2} \) the integral becomes


\(\displaystyle \displaystyle
I = \int tan(t) {dt} = \int \frac{sin(t)}{cos(t)} {dt} = -\int \frac{dcos(t)}{cos(t)} = - ln|(cos(t)| + C
\)


Substitution of t gives \(\displaystyle \displaystyle I = - ln |cos(\arctan(\frac{2x}{\sqrt3}))| = - ln | \frac{1}{ \sqrt( 1 + \frac{4x^2}{3 })} | \)


But my textbook says it's \(\displaystyle \displaystyle


\frac{ln( x^2+3/4) }{2}
\)




Thank you!

Your answer is correct (the actual work has some brain-farts lurking, but you corrected them), but this would be much easier, just substituting \(\displaystyle t = x^2+\frac{3}{4}\), and you'll have


\(\displaystyle \displaystyle \int\dfrac{\frac{1}{2} dt}{t}\).


To see that the answer is equivalent:


\(\displaystyle \displaystyle - \ln | \frac{1}{ \sqrt{ 1 + \frac{4x^2}{3 }}} | = \ln (\sqrt{1 + \frac{4x^2}{3 }} ) = \dfrac{1}{2}\ln ( 1 + \frac{4x^2}{3 } )\)

\(\displaystyle \displaystyle = \dfrac{1}{2}\ln ( \frac{4}{3}\left(\frac{3}{4} + x^2\right) ) = \dfrac{1}{2}\ln(\frac{4}{3}) + \dfrac{1}{2}\ln \left( \frac{3}{4} + x^2\right) \)

 

[suzanka]

OMG, that substitution makes it all so simple. Thank you!

And thanks for reminding me about logarithms, I've obviously forgotten some properties.