Question

Integral of [x+3/√9-x²] dx?

Answer 1

#=>I=sqrt(9-x^2)+3sin^-1(x/3)+C

Explanation:

Your Question:

`int[x+3/√9-x²] dx`
I have noted that you are trying to ask question several times.

After observing your questions ,I think that your question is:

`color(blue)(I=int(x+3)/sqrt(9-x^2)dx`

`=>I=intx/sqrt(9-x^2)dx+int3/sqrt(9-x^2)dx`

`=>I=I_1+3int1/sqrt(3^2-x^2)dx`

`=>I=I_1+3sin^-1(x/3)+c'...to(A)`

Where, `I_1=intx/sqrt (9-x^2)dx`

Take, `sqrt(9-x^2)=u=>9-x^2=u^2=>x^2=9-u^2`

`=>2xdx=2udu=>xdx=udu`

So,

`I_1=intu/udu=int1du=u+c=sqrt(9-x^2)+c`

Substituting `I_1` in `(A)`,we get

`:.I=sqrt(9-x^2)+c+3sin^-1(x/3)+c'`

`=>I=sqrt(9-x^2)+3sin^-1(x/3)+C,where, C=c+c'`