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

See below.

It has already factored the denominator.

Divide the result by the left-hand side's denominator:

We substitute these values as the integrand and add them back into our partial fractions.

In theory, you should divide the integral into smaller parts and use a substitution before integrating.

Assimilate.

Replace once more.

You can also write the final solution as follows using the logarithm's properties:

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To integrate ( \frac{3}{{(1 + x)(1 - 2x)}} ) using partial fractions, you first express the fraction in terms of partial fractions as follows:

( \frac{3}{{(1 + x)(1 - 2x)}} = \frac{A}{1 + x} + \frac{B}{1 - 2x} )

Next, you find the values of A and B by multiplying both sides by ( (1 + x)(1 - 2x) ) and simplifying:

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

Solve for A and B:

( 3 = A - 2Ax + B + Bx )

Match coefficients of x terms and constant terms:

For x term: ( -2A + B = 0 ) For constant term: ( A + B = 3 )

Solve the system of equations to find A and B:

( A = \frac{3}{3} = 1 ) ( B = 3 - A = 3 - 1 = 2 )

Now integrate each term separately:

( \int \frac{1}{1 + x} , dx = \ln|1 + x| + C_1 ) ( \int \frac{2}{1 - 2x} , dx = -\frac{1}{2} \ln|1 - 2x| + C_2 )

Combine the results:

( \int \frac{3}{{(1 + x)(1 - 2x)}} , dx = \ln|1 + x| - \frac{1}{2} \ln|1 - 2x| + C )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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