How do you integrate $int(x+1)/((x^2+4)(3x-5))$ using partial fractions?

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Answer 1 · 2018-06-26T10:17:59

$I=1/61{8ln|3x-5|-4ln|x^2+4|+7/2 tan^-1(x/2)}+c$

Here,

$I=int(x+1)/((x^2+4)(3x-5))dx$

To obtain Partial Fractions:

$color(blue)((x+1)/((x^2+4)(3x-5))=A/(3x-5)+(Bx+C)/(x^2+4)$

$=>x+1=A(x^2+4)+B(3x^2-5x)+C(3x-5)$

$=>x+1=x^2(A+3B)+x(-5B+3C)+(4A-5C)$

comparing coefficients of $x^2 ,x and$ constant term :

$A+3B=0 =>A=-3Bto(1) ,$

$-5B+3C=1=>3C=1+5B=>C=(1+5B)/3to(2)$ ,

$4A-5C=1to(3)$

Putting value of $A and C$ into $(3)$

$4(-3B)-5((1+5B)/3)=1=>-36B-5-25B=3$

$:. color(blue)(B=-8/61$

From $(1)$ ,we get , $A=-3(-8/61)=>color(blue)(A=24/61$

From $(2)$ , we get $C=(1+5(-8/61))/3=(61-40)/(3*61)=>color(blue)(C=7/61$

So,

$I=1/61int{24/(3x-5)+(-8x+7)/(x^2+4)}dx$

$=1/61{8int3/(3x-5)dx-4int(2x)/(x^2+4)dx+7int1/(x^2+2^2)dx}$

$=1/61{8ln|3x-5|-4ln|x^2+4|+7*1/2 tan^-1(x/2)}+c$

Hence,

$I=1/61{8ln|3x-5|-4ln|x^2+4|+7/2 tan^-1(x/2)}+c$

How do you integrate int(x+1)/((x^2+4)(3x-5))int(x+1)/((x^2+4)(3x-5)) using partial fractions?