-w8.3.9 int 1/(x^2 (1+x^2)) dx

[karush]

Whitman 8.3.9
$$\displaystyle
\int\frac{1 }{{x}^{2}\left(1+{x}^{2}\right)} \ dx
=-\arctan\left({x}\right)-\frac{1}{x}+C $$
Expand
$$\displaystyle
\int\frac{1}{{x}^{2}}\ dx
-\int \frac{1}{{x}^{2}+1}dx $$

Solving
$$\displaystyle
\int\frac{1}{{x}^{2}}\ dx =-\frac{1}{x}+C$$
Solving
$$x=\tan\left({u}\right) \ \ \ \ dx=\sec^2 \left({u}\right)\ du $$
$$\displaystyle -\int \frac{1}{{x}^{2}+1}dx
=-\int \frac{1}{\tan^2{u} +1}\sec^2 \left({u}\right)\ du
=\int 1 \ du
=u=-\arctan\left({x}\right)$$
Then...
$$\displaystyle -\arctan\left({x}\right)-\frac{1}{x}+C $$

 

[Greg]

You dropped a minus sign. Other than that it looks good.

 

[Prove It]

Whitman 8.3.9
$$\displaystyle
\int\frac{1 }{{x}^{2}\left(1+{x}^{2}\right)} \ dx
=-\arctan\left({x}\right)-\frac{1}{x}+C $$
Expand
$$\displaystyle
\int\frac{1}{{x}^{2}}\ dx
-\int \frac{1}{{x}^{2}+1}dx $$

I'm assuming you used Partial Fractions to do this...

Solving
$$\displaystyle
\int\frac{1}{{x}^{2}}\ dx =-\frac{1}{x}+C$$
Solving
$$x=\tan\left({u}\right) \ \ \ \ dx=\sec^2 \left({u}\right)\ du $$
$$\displaystyle -\int \frac{1}{{x}^{2}+1}dx
=-\int \frac{1}{\tan^2{u} +1}\sec^2 \left({u}\right)\ du
=\int 1 \ du
=u=-\arctan\left({x}\right)$$
Then...
$$\displaystyle -\arctan\left({x}\right)-\frac{1}{x}+C $$

This is correct, well done :)

 

[karush]

Actually I used the expand function on the TI-Nspire