# Math Help - Integration

1. ## Integration

im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

1. S p^5 ln p dp

i got ln p(1/6p^6) - 37/100 p^12 + C, but i think that is wrong

2. S 1 (x^2 + 1) e^-x dx
0

i got 1+ 3e^-1 + C

3. S b =sqrt(3) arctan (1/x) dx
a = 1

4. S b = (1) (r^3)/sqrt (4+r^2)) dr
a = 0

2. Originally Posted by skabani
im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

1. S p^5 ln p dp
$\int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx$
Let $t=\ln x$.

3. I think the answer for question 2 should be $-1 - 2e^{-1}$.

4. did u go through integration through seperation twice? or did you use a different method/

5. Originally Posted by skabani
did u go through integration through seperation twice? or did you use a different method/
First I broke the integral apart: $\int _0 ^1 x^2e^{-x} + e^{-x} dx = \int _0 ^1 x^2e^{-x} dx + \int _0 ^1 e^{-x} dx$

Then I used integration by parts twice on the first integral to yield $-x^2e^{-x} - 2xe^{-x} -2 \int _0 ^1 e^{-x} dx$.

Adding the second integral gives $-x^2e^{-x} - 2xe^{-x} - \int _0 ^1 e^{-x} dx = -x^2e^{-x} - 2xe^{-x} + e^{-x}$ (evaulated from 0 to 1).

6. thanks.

do you happen to know what to do for number 4, because i am completely stuck there...

7. Try using a trig substitution. I solved the problem using $r = 2 \tan \theta$.

8. Originally Posted by skabani
im not sure how to do these, i tried integration by parts but just got stuck or got an unsure answer.. (S= integral)

...
[snip]

4. S b = (1) (r^3)/sqrt (4+r^2)) dr
a = 0
Make the trig substitution $r=2\tan\vartheta\implies \,dr=2\sec^2\vartheta\,d\vartheta$

Thus, the integral becomes $\int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\frac{16\tan^3\vartheta \cdot\sec^2\vartheta\,d\vartheta}{4\sec\vartheta}= 4\int_0^{\tan^{-1}\left(\frac{1}{2}\right)}\tan^3\vartheta\sec\var theta\,d\vartheta$

(Hint: to integrate this, break off a factor of $\tan\vartheta$ and then apply the identity $\tan^2\vartheta=\sec^2\vartheta-1$; then apply the substitution $z=\sec\vartheta$)

Try to take it from here

--Chris

9. Originally Posted by ThePerfectHacker
$\int x^5 \ln x dx = \int e^{6\ln x} \ln x (\ln x)' dx$
Let $t=\ln x$.

im not sure how to do this still, i keep getting different anwers every time I try it by integration by parts..can you help me out?