mony01
Anyone know how to do this integral?
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mony01
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\[\int\limits \frac{ x ^{2} +x+1}{ (x ^{2}+1)^{2} }dx\]
wio
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Play around with the fraction.
wio
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Maybe partial fraction decomposition
myininaya
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did you try a trig sub i think that would work just fine
mony01
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you think i can try sin?
myininaya
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I think you can try tan
myininaya
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the hint was the x^2+1 on bottom
myininaya
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tan^2(theta)+1=sec^2(theta)
mony01
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would the set up be integral x^2+x+1/sex^2 theta
myininaya
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well you have to replace all the x's and the dx
myininaya
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and also you are leaving off the square on bottom
myininaya
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\[x=\tan(\theta) => dx=\sec^2(\theta) d \theta \]
Replace all the x's with tan(theta)
Replace the dx with sec^2(theta) d theta
mony01
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is it integral tan^2theta+tan theta+1/sec^2 theta d (theta)
myininaya
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If I think what you wrote is what I think then yes.
You mean the following I assume:
integral of (tan^2(theta)+tan(theta)+1)/sec^2(theta) d(theta)
myininaya
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\[\int\limits_{}^{}\frac{\tan^2(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{}\frac{\tan(\theta)}{\sec^2(\theta)} d \theta +\int\limits_{}^{} \frac{1}{\sec^2(\theta) }d \theta \]
myininaya
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Look at them three separately
gorv
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\[\int\limits_{}^{}(\frac{ x^2+1 }{ (x^2+1)^2 } +\frac{ x }{ (x^2+1)^2 })*dx\]
gorv
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\[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 }\]
gorv
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first one is standard formula
mony01
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how can i figure out the answer?
gorv
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\[\int\limits_{}^{}\frac{ x }{ (x^2+1)^2 } *dx\]
for this
x^2+1=t
2xdx=dt
xdx=dt/2
gorv
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\[\int\limits_{}^{} \frac{ dt }{ 2*t^2 }\]
gorv
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\[\int\limits_{}^{}\frac{ dx }{ x^2+1 } +\int\limits_{}^{}\frac{ dt }{ 2t^2 }\]
gorv
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can u solve it @mony01
mony01
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is it sec^2 (theta)/2(x^2+1)^2
myininaya
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I do like gorv's way.
But either way is fine.
gorv's is simpler though.
gorv
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actual its less calculative...both ways anns will be same
mony01
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how can i solve the rest of it?
myininaya
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What question on what part do you have?
mony01
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\[\int\limits \frac{ dx }{ x ^{2}+1}+\int\limits \frac{ dt }{ 2t ^{2} }\]
myininaya
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the first integral is just remembering that it is acran(x)
the second one 1/2 is a constant multiple and you should know how to integrate t^(-2) at this point in calculus
mony01
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is the answer arctan(x)-1/2(x^2+1)+C
myininaya
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Yes that's right you mean arctan(x)- 1/[2(x^2+1)] +C
good job