How do you find the integral of $int (4x+3)/sqrt(1-x^2)$ ?

Question

3075 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-03T06:34:39

The answer is $=-4sqrt(1-x^2)+3arcsinx+C$

We need

$intx^ndx=x^(n+1)/(n+1)+C (n!=-1)$

$sin^2x+cos^2x=1$

Rewrite the integral (we apply linearity)

$int((4x+3)dx)/sqrt(1-x^2)=int(4xdx)/sqrt(1-x^2)+int(3dx)/sqrt(1-x^2)$

$=4int(xdx)/sqrt(1-x^2)+3intdx/sqrt(1-x^2)$

Let $u=1-x^2$ , $=>$ , $du=-2xdx$

$int(xdx)/sqrt(1-x^2)=-1/2int(du)/sqrtu$

$=-1/2sqrtu/(1/2)=-sqrtu=-sqrt(1-x^2)$

Let $x=sin theta$ , $=>$ , $dx=costheta(d theta)$

$sqrt(1-x^2)=sqrt(1-sin^2theta)=sqrt (cos^2theta)=costheta$

$int(dx)/sqrt(1-x^2)=int(costheta d theta)/costheta=intd theta=theta$

$=arcsinx$

Putting it all together

$int((4x+3)dx)/sqrt(1-x^2)=-4sqrt(1-x^2)+3arcsinx+C$

How do you find the integral of int (4x+3)/sqrt(1-x^2)int (4x+3)/sqrt(1-x^2)?