How do you evaluate #int dx/(x^2-2x+2)# from #[0, 2]#?

Question

Emilia Aguilar · Accepted Answer

#pi/2#

The denominator cannot be factored, so complete the square and see what you can do in the way of u-substitution.

#int_0^2 1/(1(x^2 - 2x + 1 - 1) + 2)dx#

#int_0^2 1/(1(x^2 - 2x + 1) - 1 + 2)dx#

#int_0^2 1/((x -1)^2 + 1)dx#

Now let #u = x - 1#. Then #du = dx#. We also adjust our bounds of integration accordingly.

#int_-1^1 1/(u^2 + 1) du#

This is a standard integral.

#int_-1^1 arctan(u)#

#int_0^2 arctan(x- 1)#

Evaluate using the 2nd fundamental theorem of calculus.

#arctan(2 - 1) - arctan(0 - 1) = arctan(1) - arctan(-1) = pi/4 - (-pi/4) = pi/2#

Hopefully this helps!

Emilia Aguilar · Answer

undefined To evaluate the integral ∫dx/(x^2 - 2x + 2) from 0 to 2, you can use the method of partial fraction decomposition followed by integration. First, factor the denominator x^2 - 2x + 2 as (x - 1)^2 + 1. Then, express the integrand as a sum of two fractions with unknown constants A and B:

1 / ((x - 1)^2 + 1) = A / (x - 1) + B / ((x - 1)^2 + 1)

Solve for A and B by finding a common denominator and equating coefficients of like terms:

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

Once you have found the values of A and B, you can integrate each term separately. The integral of A / (x - 1) can be evaluated using the natural logarithm function, and the integral of B / ((x - 1)^2 + 1) can be evaluated using the arctangent function.

Finally, evaluate the definite integral from 0 to 2 using the fundamental theorem of calculus. Substitute the upper and lower limits into the antiderivatives of the fractions and subtract the result of the lower limit from the result of the upper limit.