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

Question

Paisley Johnson · Accepted Answer

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#

Charles Anctil · Answer

undefined To find the integral of $\int \frac{4x+3}{\sqrt{1-x^2}}$ with respect to $x$, you can use the substitution method. Let $u = 1 - x^2$, then $du = -2x \, dx$. Rearrange to solve for $dx$, giving $dx = -\frac{du}{2x}$. Substitute these expressions into the integral. You will end up with $\int \frac{4x+3}{\sqrt{1-x^2}} \, dx = \int \frac{4x+3}{\sqrt{u}} \cdot \left(-\frac{du}{2x}\right)$. Simplify the expression, integrate with respect to $u$, then substitute back $u = 1 - x^2$ to find the final answer.