How do you find $\int \frac{2 x}{(1 - x) (1 + x^{2})} d x$ $int (2x)/((1-x)(1+x^2)) dx$ using partial fractions?

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How do you find $\int \frac{2 x}{(1 - x) (1 + x^{2})} d x$ $int (2x)/((1-x)(1+x^2)) dx$ using partial fractions?

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Answer 1 · 2017-01-09T16:12:23

$\int \frac{2 x d x}{(1 - x) (1 + x^{2})} = - ln | 1 - x | + \frac{1}{2} ln (1 + x^{2}) - arctan x$ $int (2xdx)/((1-x)(1+x^2)) =-ln abs(1-x)+1/2ln(1+x^2)-arctanx$

Explanation:

Develop the integral in partial fractions:

$\frac{2 x}{(1 - x) (1 + x^{2})} = \frac{A}{1 - x} + \frac{B x + C}{1 + x^{2}}$ $(2x)/((1-x)(1+x^2)) = A/(1-x) + (Bx+C)/(1+x^2)$

$\frac{2 x}{(1 - x) (1 + x^{2})} = \frac{A (1 + x^{2}) + (B x + C) (1 - x)}{(1 - x) (1 + x^{2})}$ $(2x)/((1-x)(1+x^2)) = (A(1+x^2) + (Bx+C)(1-x))/((1-x)(1+x^2))$

$2 x = A + A x^{2} + B x + C - B x^{2} - C x$ $2x = A+Ax^2 +Bx+C-Bx^2-Cx$

$2 x = (A - B) x^{2} + (B - C) x + (A + C)$ $2x = (A-B)x^2 +(B-C)x+(A+C)$

So:

$A - B = 0 \Rightarrow A = B$ $A-B = 0 => A=B$

$B - C = 2 \Rightarrow A - C = 2$ $B-C = 2 => A-C=2$

$[(A-C=2) , (A+C = 0)] => A=1, B=1, C=-1$

$(2x)/((1-x)(1+x^2)) = 1/(1-x) +(x-1)/(1+x^2)$

$int (2xdx)/((1-x)(1+x^2)) = int (dx)/(1-x) +int (xdx)/(1+x^2)-int (dx)/(1+x^2)=-ln abs(1-x)+1/2ln(1+x^2)-arctanx$

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How do you find $\int \frac{2 x}{(1 - x) (1 + x^{2})} d x$ $int (2x)/((1-x)(1+x^2)) dx$ using partial fractions?

1 Answer

Explanation:

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How do you find $\int \frac{2 x}{(1 - x) (1 + x^{2})} d x$ $int (2x)/((1-x)(1+x^2)) dx$ using partial fractions?