How do you integrate $f(x)=(x^2+x)/((3x^2+2)(x+7))$ using partial fractions?

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Answer 1 · 2017-11-20T06:19:37

$int(x^2+x)/((3x^2+2)(x+7))dx$

= $23/6ln|3x^2+2|-2sqrt6tan^(-1)((sqrt3x)/(sqrt2))+ln|x+7|+c$

Let $(x^2+x)/((3x^2+2)(x+7))=(Ax+B)/(3x^2+2)+C/(x+7)$

Hence $x^2+x=(Ax+B)(x+7)+C(3x^2+2)$

or $x^2+x=Ax^2+7Ax+Bx+7B+3Cx^2+2C$

and comparing coefficients on both sides

$A+3C=1$ , $7A+B=1$ and $7B+2C=0$

the solution for these simiultaneous equations is

$A=23/149$ , $B=-12/149$ and $C=42/149$

Hence $int(x^2+x)/((3x^2+2)(x+7))dx$

= $int[1/149{(23x-12)/(3x^2+2)+42/(x+7)}]dx$

= $1/149int(23x-12)/(3x^2+2)dx+42/149int1/(x+7)dx$

We know $int1/(x+7)dx=ln|x+7|$ and also $int1/(x^2+a^2)dx=1/atan^(-1)(x/a)+c$ .

For $int(23x-12)/(3x^2+2)dx$ let us assume $u=3x^2+2$ and then $du=6xdx$

hence $int(23x-12)/(3x^2+2)dx=23/6int(6x)/(3x^2+2)dx-4int1/(x^2+2/3)dx$

= $23/6int(du)/u-4sqrt(3/2)tan^(-1)(x/(sqrt(2/3)))$

= $23/6ln|3x^2+2|-2sqrt6tan^(-1)((sqrt3x)/(sqrt2))$

Hence $int(x^2+x)/((3x^2+2)(x+7))dx$

= $23/6ln|3x^2+2|-2sqrt6tan^(-1)((sqrt3x)/(sqrt2))+ln|x+7|+c$

How do you integrate f(x)=(x^2+x)/((3x^2+2)(x+7))f(x)=(x^2+x)/((3x^2+2)(x+7)) using partial fractions?