Partial Fractions Homework Solutions
Hint. Let u = 2x − 1.
Let u = 2x − 1. Then du = 2 dx, and we get
Moral. You can integrate anything that looks like !
Hint. Let u = x2 − 5x.
Let u = x2 − 5x. Then du = 2x − 5 dx, and we get
Moral. Always check to see if you can use u-substitution before trying anything fancy!
Hint. Split up the fraction, then use
u-substitution (with u = x2 + 5) on one term and the
following formula on the other:
For the first term, let u = x2 + 5. Then du = 2x dx, and
the integral becomes
2 ln |u|+C = 2 ln |x2+5|+C. For the second term, use the formula with . So the answer
to the original integral becomes
Moral: You can integrate anything that looks like
Hint. Complete the square in the denominator, i.e.
x2+6x+11 = x2+6x+9+2 = (x+3)2+2.
Then let u = x + 3, and apply the technique in problem 3, above.
If u = x + 3 then x = u − 3. du = dx, so completing the square as in the hint, we have
Now this looks just like the previous problem. Use
u-substitution (or choose a different letter,
since we’re already in u) for the first term, and the inverse-tangent formula for the second; we get
Moral: You can integrate anything that looks like !
Perform polynomial division.
Check to make sure you get . Integrate.
We verified this long division in class. Now we have
Moral. When the integrand is an improper rational
function, perform polynomial division to
rewrite the quotient as a polynomial plus a proper rational function, then apply the previous