How do you integrate int ((x^3+1)/(x^2+3))∫(x3+1x2+3) using partial fractions?

1 Answer
maganbhai P.
Mar 17, 2018

I=x^2/2-3/2ln|x^2+3|+1/(sqrt(3))tan^-1(x/sqrt(3))+cI=x22−32ln∣∣x2+3∣∣+1√3tan−1(x√3)+c

Explanation:

Here,
(x^3+1)/(x^2+3)=(x^3+3x-3x+1)/(x^2+3)=(x^3+3x)/(x^2+3)-(3x)/(x^2+3)+1/(x^2+3)x3+1x2+3=x3+3x−3x+1x2+3=x3+3xx2+3−3xx2+3+1x2+3
(x^3+1)/(x^2+3)=(x(x^2+3))/(x^2+3)-3/2*(2x)/(x^2+3)+1/(x^2+3)x3+1x2+3=x(x2+3)x2+3−32⋅2xx2+3+1x2+3
I=int(x^3+1)/(x^2+3)dxI=∫x3+1x2+3dx
I=intxdx-3/2int(2x)/(x^2+3)dx+int1/(x^2+3)dxI=∫xdx−32∫2xx2+3dx+∫1x2+3dx
I=x^2/2-3/2int(d/(dx)(x^2+3))/(x^2+3)dx+int1/(x^2+(sqrt(3))^2)dxI=x22−32∫ddx(x2+3)x2+3dx+∫1x2+(√3)2dx
I=x^2/2-3/2ln|x^2+3|+1/(sqrt(3))tan^-1(x/sqrt(3))+cI=x22−32ln∣∣x2+3∣∣+1√3tan−1(x√3)+c

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