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

1 Answer
Aug 5, 2015

#-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#