Question #28fed

1 Answer
May 29, 2016

-1/2ln(2x^2-2x+5)+C

Explanation:

You wish to find:

int(1-2x)/(2x^2-2x+5)dx

Let u=2x^2-2x+5. This implies that du=(4x-2)dx.

Multiply the numerator of the integrand by -2. Balance this by multiplying the exterior of the integral by -1//2.

=-1/2int(-2(1-2x))/(2x^2-2x+5)dx=-1/2int(4x-2)/(2x^2-2x+5)dx

Substitute in u=2x^2-2x+5 and du=(4x-2)dx into the integrand:

=-1/2int(du)/u

Note that int(du)/u=lnabsu+C:

=-1/2ln(2x^2-2x+5)+C

Note that the absolute value signs are not needed because 2x^2-2x+5>0 for all values of x.