Ask
your own question, for FREE!
Mathematics
45 Online
OpenStudy (anonymous):
Integral of 1/(u^3 + 3u) du, without using partial fractions? Many thanks!
Still Need Help?
Join the QuestionCove community and study together with friends!
OpenStudy (anonymous):
This can be done by some none obvious algebra manipulation: integral [ 1/(u^3 + 3u) du ] = 1/3 * integral [ 3/(u^3 + 3u) du ] =1/3 * integral [ 3/{u*(u^2+3) du ] =1/3 * integral [ {3+u^2-u^2} / {u*(u^2+3)} du ] =1/3*integral [ {3+u^2} / {u*(u^2+3)} du ] - 1/3*integral [ {u^2} / {u*(u^2+3)} du ] =1/3*integral [ 1/u du ] - 1/3*integral [ u/{u^2+3} du] let u^2+3 = t => dt = 1/(2*u) du =1/3*ln(u) - 1/3 * 1/2 * integral (1/t dt) =1/3*ln(u) - 1/3*1/2*ln(t) =1/3*ln(u) - 1/3*ln(sqrt(t)) =1/3*ln( u / sqrt(t) ) =1/3*ln( u / sqrt(u^2+3) ) + C
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
XShawtyX:
Omga Help, So tmr I have a test for Math The Function Transformation, and I literally have no clue how to do any of that.
7 seconds ago
2 Replies
0 Medals
Bones:
This is to my ex best friend I am sorry for what I did I know I missed up I get t
3 hours ago
2 Replies
3 Medals
Alexis1415:
A little Italian thing I created ud83eudd26u2640ufe0f non importa quello che facc
2 hours ago
18 Replies
4 Medals