Integrate $\int\frac{1}{x(x^2+1)}dx$ - Techniques & Help

[NotaMathPerson]

I used substitution but did not get a form I know how to integrate. What technique should I use here?
$\int\frac{1}{x(x^2+1)}dx$

Thanks!

 

[Greg]

Perform the substitution

$$\tan\theta=x,\quad\sec^2\theta\,d\theta=dx$$

Can you continue with that?

 

[NotaMathPerson]

Perform the substitution

$$\tan\theta=x\quad\sec^2\theta\,d\theta=dx$$

Can you continue with that?

$$\int\frac{\sec^2\theta\,d\theta}{(\tan\theta)(1+\tan^2\theta)}$$

$$\int\frac{\sec^2\theta\,d\theta}{(\tan\theta)(\sec^2\theta)}$$

$$\int\frac{d\theta}{\tan\theta} = \int\frac{\cos\theta\,d\theta}{\sin\theta}$$

Using u substitution

$$\ln|\sin\theta| +c$$ since $$x = \tan\theta = \frac{\sin\theta}{\cos\theta}$$

$$\ln|x\cos\theta| +c$$ is this correct?

 

[Greg]

You're correct up to $\ln|\sin\theta|+C$. You need to backsub correctly, so

$$x=\tan\theta,\quad\theta=\tan^{-1}(x)$$

$$\ln|\sin(\tan^{-1}(x))|+C=\ln\left|\dfrac{x}{\sqrt{1+x^2}}\right|+C$$

See this page on inverse trig functions for information on the conversion.

 

[NotaMathPerson]

Can we integrate it using partial fractions?

 

[HallsofIvy]

Can we integrate it using partial fractions?

Of course you can- and it is pretty straight forward. Did you try?

We want to write $$\frac{1}{x(x^2+ 1)}= \frac{A}{x}+ \frac{Bx+ C}{x^2+ 1}$$.

Eliminate the fractions by multiplying both sides by $$x(x^2+ 1)$$:
$$1= A(x^2+ 1)+ (Bx+ C)x= Ax^2+ A+ Bx^2+ Cx= (A+ B)x^2+ Cx+ A$$.

So, equating "like coefficints", we must have A+ B= 0, C= 0, A= 1. A+ B= 1+ B= 0 so B= -1.

$$\int\frac{1}{x(x^2+ 1)}dx= \int \frac{1}{x} dx+ \int\frac{-x}{x^2+ 1}dx$$.

The first integral is ln|x|. To integrate the second, let $$u= x^2+ 1$$.

- - - Updated - - -

$$\int\frac{\sec^2\theta\,d\theta}{(\tan\theta)(1+\tan^2\theta)}$$

$$\int\frac{\sec^2\theta\,d\theta}{(\tan\theta)(\sec^2\theta)}$$

$$\int\frac{d\theta}{\tan\theta} = \int\frac{\cos\theta\,d\theta}{\sin\theta}$$

Using u substitution

$$\ln|\sin\theta| +c$$ since $$x = \tan\theta = \frac{\sin\theta}{\cos\theta}$$

$$\ln|x\cos\theta| +c$$ is this correct?

No, of course not. There was no "$$\theta$$" in the original integral.

 

[alyafey22]

Note that

$$\int\frac{1}{x(x^2+1)}dx=\frac {1}{2}\int\frac{\frac {-2}{x^3}}{1+\frac {1}{x^2}}dx$$

This approach can be extended to
$$\int\frac{1}{x^p(x^{p+1}+1)}dx$$