Question

How do you find the integral of `(2x-3)/((9-x^2)^0.5)dx`?

Answer 1

`-2sqrt(9-x^2)-3arcsin(x/3)+C`

Explanation:

Start off by rewriting as follows

`int (2x)/(9-x^2)^(1/2)dx-int 3/(9-x^2)^(1/2)dx`

For the first integral we use a u substitution

Let `u=9-x^2` then `(du)/dx=-2x`

`dx=(du)/(-2x)`

Make the substitution

`-int (2x)/u^(1/2)(du)/(2x)=-int u^(-1/2)du`

Integrating we have

`-2u^(1/2)`

Back substituting for u

`-2(9-x^2)^(1/2)`

`-2sqrt(9-x^2)`

For the second integral we will use a trigonometric substitution

Let `x=3sintheta` the

`dx=3costhetad theta`

Make the substitution

`int 3/(9-9sin^2theta)^(1/2)3cosd theta`

`int (9costheta)/(9[1-sin^2theta])^(1/2)d theta`

`3int costheta/(cos^2theta)^(1/2)d theta`

`3int costheta/costhetad theta`

`3int d theta`

Integrating we get `theta`

Now `theta=arcsin(x/3)`

So we have `3arcsin(x/3)`

Putting the two results together we have

`-2sqrt(9-x^2)-3arcsin(x/3)+C`