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Integration by parts - The answers proposed don't match the answers I'm getting.

Integration by parts - The answers proposed don't match the answers I'm getting.-example-1

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\displaystyle\int_{(1)/(√(3))}^(1)~tan^(-1)\left( \cfrac{1}{x} \right)dx \\\\[-0.35em] ~\dotfill\\\\ u=tan^(-1)\left( \cfrac{1}{x} \right)\implies \cfrac{du}{dx}=-\cfrac{1}{1+x^2}~\hfill v=\displaystyle\int~1\cdot dx\implies v=x \\\\[-0.35em] ~\dotfill\\\\ xtan^(-1)\left( \cfrac{1}{x} \right)+\displaystyle\int ~ \cfrac{1}{1+x^2}\cdot x\qquad \leftarrow \textit{let's do integration by parts here} \\\\[-0.35em] \rule{34em}{0.25pt}


u_1=x\implies \cfrac{du_1}{dx}=1~\hfill v_1=\displaystyle\int~\cfrac{1}{1+x^2}\implies v_1=tan^(-1)(x) \\\\\\ xtan^(-1)(x)-\displaystyle\int~tan^(-1)(x)\implies \underline{xtan^(-1)(x)-xtan^(-1)(x)}-\ln(√(1+x^2)) \\\\\\ \ln(√(1+x^2))\qquad \leftarrow \textit{now let's put it together with the left-side part} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \left. xtan^(-1)\left( \cfrac{1}{x} \right) \right]_{(1)/(√(3))}^(1) ~~-~~\left.\cfrac{}{} \ln(√(1+x^2)) \right]_{(1)/(√(3))}^(1)


~\dotfill\\\\ \stackrel{\textit{let's do firstly the left-hand-side}}{[1\cdot tan^(-1)(1)]-\cfrac{1}{√(3)}\cdot tan\left( \cfrac{~~ 1~~}{(1)/(√(3))} \right)\implies} tan^(-1)(1)-\cfrac{1}{√(3)}\cdot tan^(-1)(√(3)) \\\\\\ \cfrac{\pi }{4}-\cfrac{1}{√(3)}\cdot \cfrac{\pi }{3}\implies \cfrac{\pi }{4}-\cfrac{\pi }{3√(3)}\implies \cfrac{\pi }{4}-\cfrac{\pi }{3√(3)}\cdot \cfrac{√(3)}{√(3)}\implies \boxed{\cfrac{\pi }{4}-\cfrac{\pi √(3)}{9}}


\stackrel{\textit{now let's do the right-hand-side}}{\ln(√(1+1))-\ln\left( \sqrt{1+(1)/(3)} \right)\implies} \ln(√(2))-\ln\left( \sqrt{(4)/(3)} \right) \stackrel{\textit{let's apply the (-1) in front}}{\implies -\ln(√(2))+\ln\left( \sqrt{(4)/(3)} \right)} \\\\\\ \ln\left( \sqrt{(4)/(3)} \right)-\ln(√(2))\implies \ln\left( (2)/(√(3)) \right)-\ln(√(2))\implies \ln\left( \cfrac{~~ (2)/(√(3))~~}{√(2)} \right)


\ln\left( \cfrac{2√(2)}{√(3)} \right)\implies \ln\left( \cfrac{√(8)}{√(3)} \right)\implies \ln\left( \sqrt{\cfrac{8}{3}} \right)\implies \ln\left[ \left( \cfrac{8}{3} \right)^{(1)/(2)} \right]\implies \boxed{\cfrac{1}{2}\ln\left( \cfrac{8}{3} \right)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\pi }{4}-\cfrac{\pi √(3)}{9}~~ + ~~\cfrac{1}{2}\ln\left( \cfrac{8}{3} \right)\implies \boxed{\left( \cfrac{1}{4}-\cfrac{√(3)}{9} \right)\pi ~~ + ~~\cfrac{1}{2}\ln\left( \cfrac{8}{3} \right)}

User Leandroico
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