Integral

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Nagisa

New South Welsh
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School: Lismore High
School Grad Year: 2013

Integral

« on: July 23, 2013, 01:05:03 pm »

0

couldnt do this one, any thoughts?

$\int \frac{4t - 6}{(t+1)(2t^2+3)} dt$

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$\int\frac{1}{x}\ dx$

#1procrastinator

Victorian
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Re: Integral

« Reply #1 on: July 23, 2013, 02:05:41 pm »

+2

use partial fractions brah

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Nagisa

New South Welsh
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Posts: 114
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School: Lismore High
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Re: Integral

« Reply #2 on: July 23, 2013, 02:22:49 pm »

0

nah bra dont work m8, esh lad

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$\int\frac{1}{x}\ dx$

Nagisa

New South Welsh
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Posts: 114
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School: Lismore High
School Grad Year: 2013

Re: Integral

« Reply #3 on: July 23, 2013, 02:28:29 pm »

0

wait...

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$\int\frac{1}{x}\ dx$

Homer

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Re: Integral

« Reply #4 on: July 23, 2013, 03:52:10 pm »

+2

$\int \frac{-2}{t+1}+\frac{4t}{2t^2+3}dt$

$-2ln\left | t+1 \right |+ln\left |2t^2+3 \right |+ccccccc$

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SocialRhubarb

Victorian
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Re: Integral

« Reply #5 on: July 23, 2013, 05:11:43 pm »

+2

Quote from: Homer on July 23, 2013, 03:52:10 pm

$\int \frac{-2}{t+1}+\frac{4t}{2t^2+3}dt$

$-2ln\left | t+1 \right |+ln\left |2t^2+3 \right |+ccccccc$

You can simplify it to $\ln\frac{2t^2+3}{(t+1)^2}+C$

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