How do you integrate #int ((x^3+1)/(x^2+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))+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)#
#(x^3+1)/(x^2+3)=(x(x^2+3))/(x^2+3)-3/2*(2x)/(x^2+3)+1/(x^2+3)#
#I=int(x^3+1)/(x^2+3)dx#
#I=intxdx-3/2int(2x)/(x^2+3)dx+int1/(x^2+3)dx#
#I=x^2/2-3/2int(d/(dx)(x^2+3))/(x^2+3)dx+int1/(x^2+(sqrt(3))^2)dx#
#I=x^2/2-3/2ln|x^2+3|+1/(sqrt(3))tan^-1(x/sqrt(3))+c#

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