How do you integrate $int (2x - 1) / (x^2 + 5x - 6) dx$ using partial fractions?

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Answer 1 · 2016-10-31T13:49:12

The integral equals $13/7ln|x + 6| + 1/7ln|x - 1| + C$ .

We can factor the denominator as $(x + 6)(x - 1)$ .

So,

$A/(x + 6) + B/(x- 1) = (2x - 1)/((x + 6)(x - 1))$

$A(x - 1) + B(x + 6) = 2x - 1$

$Ax - A + Bx + 6B = 2x - 1$

$(A + B)x +(-A + 6B) = 2x - 1$

We can now write a systems of equations to solve for $A$ and $B$ .

$A + B = 2$
$-A + 6B = -1$

$B = 2- a$

$-A + 6(2 - A) = -1$

$-A + 12 - 6A = -1$

$-7A = -13$

$A = 13/7$

$13/7 + B = 2$

$B = 2- 13/7$

$B = 1/7$

Resubstitute into the initial equation.

The integral become $int((13/7)/(x + 6) + (1/7)/(x - 1)) dx$

We now use the rule $int(1/x)(dx) = ln|x| + C$ to get rid of the integral.

Letting the function be $f(x)$ .

$f(x) = 13/7ln|x + 6| + 1/7ln|x - 1| + C$

Hopefully this helps!

How do you integrate int (2x - 1) / (x^2 + 5x - 6) dxint (2x - 1) / (x^2 + 5x - 6) dx using partial fractions?