int (ln(x^2 + 9x +1) dx help solve
C Catherine New member Joined Sep 27, 2006 Messages 1 Sep 27, 2006 #1 int (ln(x^2 + 9x +1) dx help solve
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 27, 2006 #2 Re: calculus Hello, Catherine! \(\displaystyle \L\int\)\(\displaystyle \ln(x^2\,+\,9x\,+\,1)\,dx\) Click to expand... Integrate it "by parts" . . . . . \(\displaystyle \begin{array}{ccc}u\,=\,\ln(x^2\,+\,9x+1) & \;\;\; & dv\,= \,dx \\ . \\ du\,=\,\frac{2x\,+\,9}{x^2\,+\,9x\,+\,1}\,dx & \;\;\; & v\,=\,x\end{array}\) We have: \(\displaystyle \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle x\cdot\frac{2x\,+\,9}{x^2\,+\,9x\,+\,1}\,dx\) . . . . . \(\displaystyle = \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle \frac{2x^2\,+\,9x}{x^2\,+\,9x\,+\,1}\,dx\) . . . . . \(\displaystyle = \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle \left(2\,-\,\frac{9x\,+\,2}{x^2\,+\,9x\,+\,1}\right)\,dx\) Can you finish it now/
Re: calculus Hello, Catherine! \(\displaystyle \L\int\)\(\displaystyle \ln(x^2\,+\,9x\,+\,1)\,dx\) Click to expand... Integrate it "by parts" . . . . . \(\displaystyle \begin{array}{ccc}u\,=\,\ln(x^2\,+\,9x+1) & \;\;\; & dv\,= \,dx \\ . \\ du\,=\,\frac{2x\,+\,9}{x^2\,+\,9x\,+\,1}\,dx & \;\;\; & v\,=\,x\end{array}\) We have: \(\displaystyle \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle x\cdot\frac{2x\,+\,9}{x^2\,+\,9x\,+\,1}\,dx\) . . . . . \(\displaystyle = \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle \frac{2x^2\,+\,9x}{x^2\,+\,9x\,+\,1}\,dx\) . . . . . \(\displaystyle = \;x\cdot\ln(x^2\,+\,9x\,+\,1)\,-\,\L\int\)\(\displaystyle \left(2\,-\,\frac{9x\,+\,2}{x^2\,+\,9x\,+\,1}\right)\,dx\) Can you finish it now/
