Evaluating a Definite Integral using Substitution

Hi syfa, let's solve this definite integral together using the substitution method.

We are asked to evaluate the integral from zero to one of x plus one over the quantity x squared plus two x plus six squared.

Notice that the numerator looks like the derivative of the expression inside the parentheses in the denominator. So, let's set u equal to x squared plus two x plus six.

Now, we find the derivative of u with respect to x. d u equals two x plus two, all times d x.

We can factor out a two from the right side, giving us d u equals two times the quantity x plus one times d x.

Solving for d x, we get d x equals d u over two times the quantity x plus one.

Since this is a definite integral, we must also change our limits of integration from x values to u values.

For our lower limit, when x equals zero, substituting it into our u expression gives us zero squared plus zero plus six, which equals six.

For our upper limit, when x equals one, we get one squared plus two times one plus six, which equals nine.

Now, let's rewrite the integral in terms of u using our new limits and the substitution.

The integral from zero to one becomes the integral from six to nine. We replace the denominator with u squared and d x with d u over two times x plus one.