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 December 29th, 2012, 05:37 PM #1 Newbie   Joined: Jun 2012 Posts: 15 Thanks: 0 Evaluate Integral ?(x)(In(1+X^2)dx My attempt, On the link:
 December 29th, 2012, 06:05 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Evaluate Integral I think your substitution is good: $w=x^2+1\,\therefore\,dw=2x\,dx$ and we have: $\frac{1}{2}\int\ln(w)\,dw$ Now, use integration by parts: $u=\ln(w)\,\therefore\,du=\frac{1}{w}\,dw$ $dv=dw\,\therefore\,v=w$ and we have: $\frac{1}{2}$$w\ln(w)-\int\,dw$$=\frac{1}{2}w$$\ln(w)-1$$+C=\frac{1}{2}$$x^2+1$$$$\ln\(x^2+1$$-1\)+C=\frac{1}{2}$$\(x^2+1$$\ln$$x^2+1$$-x^2\)+C$
December 29th, 2012, 06:20 PM   #3
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Re: Evaluate Integral

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 Originally Posted by MarkFL I think your substitution is good: $w=x^2+1\,\therefore\,dw=2x\,dx$ and we have: $\frac{1}{2}\int\ln(w)\,dw$ Now, use integration by parts: $u=\ln(w)\,\therefore\,du=\frac{1}{w}\,dw$ $dv=dw\,\therefore\,v=w$ and we have: $\frac{1}{2}$$w\ln(w)-\int\,dw$$=\frac{1}{2}w$$\ln(w)-1$$+C=\frac{1}{2}$$x^2+1$$$$\ln\(x^2+1$$-1\)+C=\frac{1}{2}$$\(x^2+1$$\ln$$x^2+1$$-x^2\)+C$
Mark, do we always neglect the variable and only use to the coefficient to put in front of the integral after differentiating u?

 December 29th, 2012, 06:34 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Evaluate Integral I'm not sure what you mean, but we may look at the integral as: $\frac{1}{2}\int\ln$$x^2+1$$\,2x\,dx$ Since we let $w=x^2+1\,\therefore\,dw=2x\,dx$ this means we may write: $\frac{1}{2}\int\ln$$w$$\,dw$
December 29th, 2012, 06:49 PM   #5
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Re: Evaluate Integral

Quote:
 Originally Posted by MarkFL I'm not sure what you mean, but we may look at the integral as: $\frac{1}{2}\int\ln$$x^2+1$$\,2x\,dx$ Since we let $w=x^2+1\,\therefore\,dw=2x\,dx$ this means we may write: $\frac{1}{2}\int\ln$$w$$\,dw$
And if for instance the derivative was 3/2x multiplied by dx, then it would be 3/2 behind the integral symbol. So we would basically neglect the variable?

 December 29th, 2012, 06:59 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,193 Thanks: 504 Math Focus: Calculus/ODEs Re: Evaluate Integral If we had $dw=\frac{3}{2}x\,dx$ then we would have the reciprocal in front of the integral to effectively be multiplying the integral by 1, so that we do not change its value. In the problem above we needed a 2 with the x to get the differential, so we countered with a 1/2 in front.

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