Question

Compute the following integral. Clearly show the substitution used for the following integral and how it was used.

(8y+15/36+y^2)dy

sorry i couldnt get that sign for integral that should be at the front.

Answer 1

Break it up:
\[\int\limits\frac{8y}{36+y^{2}} dy +\int\limits\frac{15}{36+y^{2}} dy\]
Then integrate it one by one.
The first one is an inverse function. And the second is a u-substitution.

Answer 2

\[\int\limits_{}^{}\frac{8y}{36+y ^{2}}dy=4\ln(36+y ^{2})+C\]
First one

Answer 3

wait..they are both trig inverse functions..my bad.

Answer 4

yes second function is an arctan

Answer 5

Second one
\[\frac{\tan ^{-1}(\frac{y}{6})}{6}+C\]

Answer 6

\[\int\limits\limits\frac{15}{36+y^{2}} dy = 15\int\limits\limits\frac{1}{6^2+y^{2}} dy = \frac{15}{6} \tan^{-1} \frac{y}{6}= \frac{5}{2} \tan^{-1} \frac{y}{6}\]
So finally

\[4\ln(36+y ^{2}) +\frac{5}{2} \tan^{-1} \frac{y}{6} +C\]

Answer 7

No here a is 6

Answer 8

a^2+ y^2 = 36 + y^2 = (6)^2 + y^2

Answer 9

\[\int\limits\frac{1}{a^{2}+x^{2}} dx = \frac{1}{a} \tan^{-1}\frac{x}{a} \]

Answer 10

\[\frac{1}{6} \tan^{-1} \frac{y}{6}\]

Answer 11

Diya, you were correct at the first place..

Answer 12

15 \[ 15\int\limits\limits\limits\frac{1}{6^2+y^{2}} dy = 15 \times \frac{1}{6} \tan^{-1} \frac{y}{6}= \frac{5}{2} \tan^{-1} \frac{y}{6} \]

Answer 13

Thank you guys. I hope i get ready for my exam. have to keep trying to understand though.

Answer 14

Goodluck :)

Answer 15

Thanks