Using partial fractions, find ∫ 8/(3x^2+1)(3x+1) dx

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`8/((3x^2+1)(3x+1)) = (Ax+B)/(3x^2+1)+C/(3x+1)`

`8 = (Ax+B)(3x+1)+C(3x^2+1)`

`8 = 3x^2(A+C)+x(3B+A)+B+C`

Comparing components at both sides;

`x^2 rarr 0 = A+C ----(1)`

`x rarr 0 = A+3B ----(2)`

`cons. rarr 8 = B+C ---(3)`

Solving (1),(2) and (3) gives;

`A = -6`

`B = 2`

`C = 6`

` 8/((3x^2+1)(3x+1)) = (Ax+B)/(3x^2+1)+C/(3x+1)`

`8/((3x^2+1)(3x+1)) = (-6x+2)/(3x^2+1)+6/(3x+1)`

`int[8/((3x^2+1)(3x+1)]dx`

`= int(-6x+2)/(3x^2+1)dx+int6/(3x+1)dx`

`= -int(6x)/(3x^2+1)dx+int2/(3x^2+1)dx+6int1/(3x+1)dx`

`= -int(6x)/(3x^2+1)dx+(2/3)int1/(x^2+1/3)dx+6int1/(3x+1)dx`

`= -ln(3x^2+1)+(2/3)xx1/sqrt(3)tan^(-1)(sqrt3x)+2ln(3x+1)`

`= ln((3x+1)^2/(3x^2+1))+(2/(sqrt(3))tan^(-1)(sqrt3x)+C`

Where C is a constant.

**Sources:**

`int (8dx)/((3x^2+1)(3x+1))` `=2 [int ((1-3x)dx)/(1+3x^2)+3int (dx)/(1+3x)]=`

`= 2 int (dx)/(1+3x^2)-int (6xdx)/(1+3x^2)+2 int (3dx)/(1+3x)=`

`=2/3 sqrt(3) int (sqrt(3)dx)/(1+(xsqrt(3))^2)` `-int (d(1+3x^2))/(1+3x^2)+` `2 int (d(1+3x))/(1+3x)=`

`2/3 sqrt(3) int (d(xsqrt(3)))/(1+(xsqrt(3))^2)` `- int (d(1+3x^2))/(1+3x^2) +2 int (d(1+3x))/(1+3x)=`

`2/3sqrt(3) arctan (xsqrt(3))` `-ln(1+3x^2)+2ln(1+3x)+c=`

`=2/3sqrt(3) arc tan (x sqrt(3)) + ln[(1+3x)^2/(1+3x^2)] +c`

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