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1.20 Two distinct linear factors

Next we consider the case where we have two distinct linear factors in the denominator.

Example.

Find

3x+3x2+x-2dx.\int{{3x+3}\over{x^{2}+x-2}}dx.

Solution. We factorize the quadratic: x2+x-2=(x+2)(x-1)x^{2}+x-2={(x+2)(x-1)} then we take partial fractions

3x+3(x+2)(x-1)=Ax+2+Bx-1 3x+3=A(x-1)+B(x+2){{{3x+3}\over{(x+2)(x-1)}}={A\over{x+2}}+{B\over{x-1}}\Rightarrow}\,{3x+3=A(x-% 1)+B(x+2)}

whence A=1,A={1,} B=2.B={2.} Now we integrate

Adxx+2+Bdxx-1=log|x+2|+2log|x-1|+c.\int{{A\,dx}\over{x+2}}+\int{{B\,dx}\over{x-1}}={\log|x+2|+2\log|x-1|+c.}