Solve by not using partial fractions.
I liked this particular integral, I solved it on another forum to show that sometimes we can avoid the annoying algebra when doing the partial fractions.
Follow Math Help Forum on Facebook and Google+
Sub. on the third integral , we have
For the second integral , use integration by parts ,
I observe that:
So we have:
Where for the last integral we use the substitution to get:
Letting in the remaining integral we have:
good! pretty different solutions from the one i have:
consider (1); now from put and the integral becomes (2).
on the original integral put then
from (1) and (2) the latter equals
and the original integral equals
What about letting right away. The integral becomes
Then backsubstitute noting that
View Tag Cloud