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sjg13e
 one year ago
Integration by Parts problem
sjg13e
 one year ago
Integration by Parts problem

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sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\tan^{1}xdx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try u = arctan x dv = dx

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1\[\let u = \tan^{1} and du = \frac{ 1 }{ x^2 +1 }dx\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1so, \[\tan^{1} \frac{ 1 }{ x } x  \int\limits_{}^{}xdx = \tan^{1} \frac{ 1 }{ x } x  .5x^2 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0* du should be positve du 1/(x²+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's the derivative of arctan x. \[\frac{ d }{ dx }\tan^{1} x=\frac{ 1 }{ 1+x^2 }\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1we need to take the derivative of the inside function, which is 1/x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. that's an inverse function, not an exponent

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1\[(1/x)\prime = 1/x^2\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1i think it's treated as an exponent

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.11/x = x^1 = 1/2 * x^2

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1ohh I'm sorry!! i forgot to properly write the initial problem. It's supposed to say \[\int\limits_{}^{}\tan^{1} (\frac{ 1 }{ x }) dx\]

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1i apologize. that's totally my fault

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1okay, ahah, so sorry about that! so besides that little mishap, does everything else look right?

sjg13e
 one year ago
Best ResponseYou've already chosen the best response.1Just for future reference. My previous answer was wrong, here's the correct, complete solution: \[\int\limits_{}^{}\tan^{1} (\frac{ 1 }{ x })dx = x \tan^{1}(\frac{ 1 }{ x })  \int\limits_{}^{}x*\frac{ 1 }{ x^2 +1 }dx \] at this point, you have to do usubsition in the integral. so, let u = x^2 + 1, which will make .5du = xdx we can bring the  and the .5 out front to give: \[x \tan^{1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\int\limits_{}^{}\frac{ du }{ u }\] \[ =x \tan^{1} (\frac{ 1 }{ x }) + \frac{ 1 }{ 2 }\lnu \] \[= x \tan^{1} (\frac{ 1 }{ x }) + \ln x^2 + 1 + C\]
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