How do you integrate ∫x−3x2(x+3)(x−1)(x−7) using partial fractions?

1 Answer
Parn W.
Sep 27, 2016

=−34In(x+3)+112In(x−1)−73In(x−7)

Explanation:

first ,ignore the integrate sign and do partial fraction of the function

x−3x2(x+3)(x−1)(x−7)=Ax+3+Bx−1+Cx−7

x−3x2=A(x−1)(x−7))+B(x+3)(x−7))+C(x+3)(x−1)

subtitute x=−3

(−3)−3(−3)2=A(−3−1)(−3−7)

−3−27=A(−4)(−10)

−30=40A

A=−34

subtitute x=1

(1)−3(1)2=B(1+3)(1−7)

−2=B(4)(−6)

−2=−24B

B=112

subtitute x=7

(7)−3(7)2=C(7+3)(7−1)

7−147=C(10)(6)

−140=60C

C=−73

∫x−3x2(x+3)(x−1)(x−7)

=∫−34(x+3)+112(x−1)−73(x−7)

=−34∫1x+3+112∫1x−1−73∫1x−7

=−34In(x+3)+112In(x−1)−73In(x−7)

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
1603 views around the world
You can reuse this answer
Creative Commons License