Question

Integrals, enjoy :)

NO WOLFRAM ALLOWED

Answer 1

where are intergrals BTW..?

Answer 2

Hmm *scratchs soap beard*

Answer 3

\[\int\limits \frac{ dt }{ t^2\sqrt{t^2-16} }\]

\[\int\limits\limits \tan^2 \theta \sec ^{4} \theta\]
\[\int\limits \frac{ x^2+2x-1 }{ x^3-x }dx\]

Answer 4

Forgot to add dt and dtheta on the first 2 lol

Answer 5

Takes me a while to type those in equation bar haha sorry!

Answer 6

hahahah !

Answer 7

Show your work!

Answer 8

Does it count if the first one is in my integral table in my book?

Answer 9

the second one :
take z=tan(theta)
dz=sec^2(theta) d(theta)
now:
\[\int\limits_{?}^{?}z^2(1+z^2)dz\]

Answer 10

Yup, can you do the others, for full marks :P

Answer 11

@random231 .... this question was for batman's enjoyment...lol

Answer 12

Testing the everyday calculus student :)

Answer 13

\[\int\limits_{0}^{\infty} \frac{ e ^{-x \sqrt{3}} }{ x }(1-sinx)(1+2sinx-\cos2x)dx\]

@hartnn This one is for you :)

Answer 14

Okay, since you said no wolfram, and I'm not using wolfram, but a table of integrals in my book, and that makes it cool:

\[\int\limits{\frac{dt}{t^2\sqrt{t^2-16}}}\]
because:
\[\int\limits{\frac{du}{u^2\sqrt{u^2-a^2}}} = \frac{\sqrt{u^2-a^2}}{a^2u} +C \]
\[= \frac{\sqrt{t^2-16}}{16t} +C\]

Answer 15

Gj :) even though it was in your book haha

Answer 16

1 point for @random231 1 point for @heril who wants to do the third one, and also if you like try the one I gave hartnn :)

Answer 17

I just have the book laying around, even though I'm done with calculus now.

Answer 18

u - sub will not work :)

Answer 19

I got the third one, just need to type it.

Answer 20

im back after lunch!!

Answer 21

By using partial fraction decomposition, we simplify it:
\[\frac{x^2+2x-1}{x^3+x} =\frac{x^2+2x-1}{x(x^2+1) } = \frac{A}{x^2+1}+\frac{B}{x}\]
\[x^2+2x-1 = Ax + B(x^2+1), B = -1,A=2x+2 \]
so:
\[\frac{x^2+2x-1}{x^3+x} = \frac{2x+2}{x^2+1}-\frac{1}{x}\]
We can then take the integral as follows:
\[ \frac{2x+2}{x^2+1} = \frac{2x}{x^2+1}+ \frac{2}{x^2+1}\]
\[\int\limits \frac{2x}{x^2+1}dx, u =x^2+1, du = 2xdx, \int\limits \frac{1}{u}du = \ln(u)+C = \ln(x^2+1) + C\]
Again by my handy table of integrals, since I don't really want to write out the trig substitution\[\int\limits \frac{2}{x^2+1}dx = 2\arctan x +C\]
Finally:
\[\int\limits -\frac{1}{x}dx = -\ln x + C\]
Bringing a final result of:
\[\ln(x^2 + 1) + \arctan x - \ln x +C\]

Answer 22

oops, final answer inconsistent with my equations:
\[\ln(x^2 + 1) + 2\arctan x - \ln x + C \]

Answer 23

mhm

Answer 24

You should've done A/x +b/(x+1)+C/(x-1) you were close, and I will give you a medal for the effort :P!

Answer 25

http://www.wolframalpha.com/input/?i=integrate+%28x%5E2%2B2x-1%29%2F%28x%5E3-x%29dx

Answer 26

Hey, I was able to split up the fraction properly anyways, so I'd say it counts.

Answer 27

Well done heril, you have won the integration nation tournament! o_O