How do you integrate $\int \frac{1}{(1 + x) (1 - 2 x)}$ $int 1/[(1+x)(1-2x)]$ using partial fractions?

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Answer 1 · 2016-03-16T13:55:13

$\int \frac{1}{(1 + x) (1 - 2 x)} d x$ $int1/((1+x)(1-2x)]dx$

= $\frac{1}{3} ln (1 + x) - \frac{1}{3} ln (1 - 2 x) + c$ $1/3ln(1+x)-1/3ln(1-2x)+c$

To integrate, we should first convert $\frac{1}{(1 + x) (1 - 2 x)}$ $1/((1+x)(1-2x))$ in to partial fractions. Let

$\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{A}{1 + x} + \frac{B}{1 - 2 x}$ $1/((1+x)(1-2x))hArrA/(1+x)+B/(1-2x)$ . Simplifying RHS

= $\frac{A (1 - 2 x) + B (1 + x)}{(1 + x) (1 - 2 x)}$ $[A(1-2x)+B(1+x)]/((1+x)(1-2x)$ or

= $\frac{A - 2 A x + B + B x}{(1 + x) (1 - 2 x)}$ $[A-2Ax+B+Bx]/((1+x)(1-2x)$

= $\frac{(B - 2 A) x + (A + B)}{(x + 3) (x - 7) (x + 4)}$ $[(B-2A)x+(A+B)]/((x+3)(x-7)(x+4))$

Hence $B - 2 A = 0$ $B-2A=0$ , $A + B = 1$ $A+B=1$ or $A = 1 - B$ $A=1-B$

Hence $B - 2 (1 - B) = 0$ $B-2(1-B)=0$ or $3 B = 2$ $3B=2$ or

$B = \frac{2}{3}$ $B=2/3$ and hence $A = \frac{1}{3}$ $A=1/3$

Hence $\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)}$ $1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))$

Hence $\int \frac{1}{(1 + x) (1 - 2 x)} d x$ $int1/((1+x)(1-2x))dx$ =

$\int [\frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)}] d x$ $int[1/(3(1+x))+2/(3(1-2x))]dx$

Now one can use the identity $\int (\frac{1}{a x + b}) d x = \frac{1}{a} ln (a x + b)$ $int(1/(ax+b))dx=1/aln(ax+b)$

Hence, $\int [\frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)}] d x$ $int[1/(3(1+x))+2/(3(1-2x))]dx$

= $\frac{1}{3} ln (1 + x) + \frac{2}{3} \times (- \frac{1}{2}) ln (1 - 2 x) + c$ $1/3ln(1+x)+2/3xx(-1/2)ln(1-2x)+c$

= $\frac{1}{3} ln (1 + x) - \frac{1}{3} ln (1 - 2 x) + c$ $1/3ln(1+x)-1/3ln(1-2x)+c$

How do you integrate ∫1(1+x)(1−2x)int 1/[(1+x)(1-2x)] using partial fractions?