How do you evaluate this integral?

Question

What is the integral of
dx/(x+1)(x^2+1)? The numerator is dx and the rest is the denominator. Use the method of partial fractions to solve this integral.
Also, what is the integral of (e^tdt)/ (e^2t+3e^t+2)?

Answer 1

For partial fractions, you split the denominator into it's 2 component fractions, and then name the numerators A and B. In this case, you use Bx + C since it's an irreducable quadratic equation:

1 / (x+1)(x^2+1) = A / (x + 1) + (Bx + C)/(x^2 + 1)

Multiply both sides by (x+1)(x^2+1):
(x+1)(x^2+1) / (x+1)(x^2+1) = A(x+1)(x^2+1) / (x+1) + (Bx + C)(x+1)(x^2+1) / (x^2 + 1)
1 = A(x^2 + 1) + (Bx + C)(x + 1)

Let x = -1, which helps us by making x + 1 = 0:
1 = A((-1)^2 + 1)
1 = A(2)
A = 1/2

We can now write it as:
1 = 1/2(x^2 + 1) + (Bx + C)(x + 1)
If x = 1, that makes the first term equal 1 again:
1 = 1/2(1 + 1) + (B + C)(1 + 1)
1 = 1 + 2(B + C)
0 = 2(B + C)
B + C = 0
B = -C

Now let x = 0 to get rid of B:
1 = A(x^2 + 1) + (Bx + C)(x + 1)
1 = A(0 + 1) + (B(0) + C)(0 + 1)
1 = A + C
1 = 1/2 + C
C = 1/2
B = -C = -1/2

Now that we've defined A, B, and C, we can go back to our original partial fraction:
A / (x + 1) + (Bx + C)/(x^2 + 1)
Substitute the values:
1 / 2(x+1) + (-1/2x + 1/2)/(x^2 + 1)

Integrate the first fraction since it's easy:
dx / 2(x+1)
u = x + 1
du = dx
integrate du / 2u = 1/2 ln |u| = 1/2 ln |x + 1| + C (part 1)

Split out the second fraction into 2 more fractions:
(-1/2x + 1/2)/(x^2 + 1)
1/2 (1 - x) dx/(x^2 + 1)
(1/2 dx / (x^2 + 1)) - (1/2 x dx / (x^2 + 1))

The second fraction above is easier to integrate:
integral:1/2 -x dx / x^2 + 1
u = x^2 + 1
du = 2x dx
1/2 du = x dx

integral:1/2 * -1/2 du / u = -1/4 ln |u| = -1/4 ln |x^2 + 1| + C (part 2)

That leaves the first fraction:
1/2 dx / (x^2 + 1)

This is best solved by reverse substituting with a trigonometric identity: tan^2 u + 1 = sec^2 u
x = tan u
dx = (sec^2 u) du

Substitute x for tan u:
1/2 (sec^2 u) du / ((tan^2 u) + 1)
1/2 (sec^2 u) du / (sec^2 u) (the sec^2 u's helpfully cancel here)
1/2 du
integral of 1/2 du: = 1/2u = 1/2 tan u + C (part 3)

Now that we've got all 3 fractions integrated, we add up the answers for our solution:
1/2 ln |x + 1| - 1/4 ln |x^2 + 1| + 1/2 tan u + C (solution)

I'll let you do the second one - especially given how much space the first one took :)