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

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Answer 1 · 2017-01-06T21:28:10

See below.

$int3/((1+x)(1-2x))=ln(|1+x|)-ln(|1-2x|)+C$

The denominator is already factored.

$=>3/((1+x)(1-2x))=A/(1+x)+B/(1-2x)$

Multiply through by the denominator of the left-hand side:

$(1+x)(1-2x)[3/(cancel((1+x)(1-2x)))=A/cancel(1+x)+B/cancel(1-2x)]$

=> $3=A(1-2x)+B(1+x)$

The next step is to solve for $A$ and $B$ . One way to do this is to pick values for $x$ which will cancel each variable.

$x=1/2$

$3=cancel(A*0)+3/2B$

$B=2$

$x=-1$

$3=A(1-(-2))+cancel(B*0)$

$3=3A$

$A=1$

We put these values back into our partial fractions and replace as the integrand.

$int(1)/(1+x)+2/(1-2x)dx$

Technically, you should use a substitution before integrating. Split up the integral.

$int1/(1+x)dx+2int1/(1-2x)dx$

For the first integral, $u=1+x, du=dx$
For the second integral, $z=1-2x, dz=-2dx=>-1/2dz=dx$

$int1/udu-int1/zdz$

Integrate.

$ln(|u|)-ln(|z|)+C$

Substitute back in.

$ln(|1+x|)-ln(|1-2x|)+C$

Note: the absolute value signs account for the domain of the natural log function ( $x>0$ ).

By the properties of logarithms, you may also write the final answer as:

$ln(|(1+x)/(1-2x)|)+C$

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