Use partial fractions to find the indefinite integral. ∫x2+12x+12x3−4xdx

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Use partial fractions to find the indefinite integral.  ∫x2+12x+12x3−4xdx

Bob Huerta · Accepted Answer

Step 1  Given: I=∫x2+12x+12x3−4xdx  For evaluating given integral, first we simplify given expression then integrate it  Step 2  So,  I=∫x2+12x+12x3−4xdx  =∫x3+12x+12x(x2−4)dx  =∫x3+12x+12x(x2−22)dx   (∵a2−b2=(a+b)(a−b))  =∫x2+12x+12x(x+2)(x−2)dx  =∫(5x−2−3x−1x+2)dx   (∵∫dxx+a=ln|x+a|+c)  =5ln|x−2|−3ln|x|−ln|x+2|+c  Hence, given integral can be find as above.

poleglit3 · Accepted Answer

∫x2+12x+12x3−4xdx x2+12x+12x3−4x=Ax+Bx−2+Cx+2 ⇒x2+12x+12=A(x2−4)+B×(x+2)+C×(x−2) put x=0 12=A(−4)+B×0+C×0 ⇒A=−3 put x=2 4+24+12=B2(2+2) ⇒40=B×8 ⇒ B=5 put x=−2,(−2)2+12(−2)+12=A((−2)2−4)+B(−2)(2−2)+C(−2)(−2−2) ⇒4−24+12=8C ⇒−8=8C⇒C=−1 x2+12x+12x3−4x=5x−2−1x+2−3x ∫x2+12x+12x3−4xdx=5∫dxx−2−∫dxx+2−3∫dxx =5ln⁡(x−2)−ln⁡(x+2)−3ln⁡x+C =ln⁡(x−2)5x3(x+2)+C

karton · Accepted Answer

Given: ∫x2+12x+12x3−4xdx∫−3x+5x−2−1x+2dx−∫3xdx+∫5x−2dx−∫1x+2dx−3ln⁡(|x|)+5ln⁡(|x−2|)−ln⁡(|x+2|)Answer:−3ln⁡(|x|)+5ln⁡(|x−2|)−ln⁡(|x+2|)+C

Use partial fractions to find the indefinite integral. \int \frac{x^{2}+12x+12}{x^{3}-4x}dx

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