Question

Integrate 9/ (4+x^2)????

I factored out a 9 from the numerator and a 4 from the denominator, but what next? thanks for the help!

Answer 1

`int9/(4+x^2)dx=9/2tan^-1(x/2)+C`

Explanation:

`int9/(4+x^2)dx=9intdx/(4+x^2)`

Factoring out a `4` from the denominator yields:

`9intdx/(4+x^2)=9intdx/(4(1+x^2/4))`

The `4` we factored out from the denominator can be brought out of the integral as `1/4`:

`(9)(1/4)intdx/(1+(x^2/4))`

`(x^2/4)=(x/2)^2`. Rewrite:

`9/4intdx/(1+(x/2)^2)`

Make the following substitution:

`u=x/2`
`(du)/dx=1/2`

`du=1/2dx`

`2du=dx`

Rewrite the integral in terms of `u,` factoring out the `2` from `du:`

`(2)(9/4)int(du)/(1+u^2)`

Recall the common integral,

`intdx/(1+x^2)=tan^-1(x)`

Therefore, `int(du)/(1+u^2)=tan^-1(u)`:

`(2)(9/4)int(du)/(1+u^2)=9/2tan^-1(u)+C`

Don't forget our added constant. Rewrite in terms of `x:`

`9/2tan^-1(u)+C=9/2tan^-1(x/2)+C`

Thus,

`int9/(4+x^2)dx=9/2tan^-1(x/2)+C`