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Integration by parts
- Thread starter serenity
- Start date
tutor_joel
Junior Member
- Joined
- Feb 6, 2010
- Messages
- 157
ok, good so far.
\(\displaystyle uv-\int{vdu}\)
I would use the tabular integration by parts for this, it looks a little long. Let me know if you need help with this method, it's actually really easy.
\(\displaystyle dv =\frac{1}{\sqrt{1+x^2}}\)
\(\displaystyle v=sinh^{-1}x\)
\(\displaystyle u=x^3\)
\(\displaystyle du=3xdx\)
\(\displaystyle x^3sinh^{-1}x-3\int{xsinh^{-1}dx\)
..and now you have to do it again...and then again...and then you'll use the tab method because it sets it all up for you.
hope this helps
\(\displaystyle uv-\int{vdu}\)
I would use the tabular integration by parts for this, it looks a little long. Let me know if you need help with this method, it's actually really easy.
\(\displaystyle dv =\frac{1}{\sqrt{1+x^2}}\)
\(\displaystyle v=sinh^{-1}x\)
\(\displaystyle u=x^3\)
\(\displaystyle du=3xdx\)
\(\displaystyle x^3sinh^{-1}x-3\int{xsinh^{-1}dx\)
..and now you have to do it again...and then again...and then you'll use the tab method because it sets it all up for you.
hope this helps
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Another \ way:\)
\(\displaystyle \int\frac{x^{3}}{(x^{2}+1)^{1/2}}dx, \ Let \ u \ = \ (x^{2}+1)^{1/2}, \ then \ x^{2} \ = \ u^{2}-1\)
\(\displaystyle and \ du \ = \ \frac{xdx}{u}, \ \implies \ dx \ = \ \frac{udu}{x}\)
\(\displaystyle Hence, \ \int\frac{x^{3}u}{xu}du \ = \ \int x^{2}du \ = \ \int(u^{2}-1)du \ = \ \frac{u^{3}}{3}-u+C\)
\(\displaystyle Resub: \ \frac{(x^{2}+1)^{3/2}}{3}-(x^{2}+1)^{1/2}+C \ = \ \frac{(x^{2}+1)^{3/2}-3(x^{2}+1)^{1/2}}{3}+C\)
\(\displaystyle = \ \frac{(x^{2}+1)^{1/2}[(x^{2}+1)-3]}{3}+C \ = \ \frac{(x^{2}+1)^{1/2}(x^{2}-2)}{3}+C\)
\(\displaystyle Check: \ D_x\bigg[\frac{(x^{2}+1)^{1/2}(x^{2}-2)}{3}+C\bigg] \ = \ \frac{x^{3}}{(x^{2}+1)^{1/2}}, \ a \ good \ exercise \ for \ you.\)
\(\displaystyle \int\frac{x^{3}}{(x^{2}+1)^{1/2}}dx, \ Let \ u \ = \ (x^{2}+1)^{1/2}, \ then \ x^{2} \ = \ u^{2}-1\)
\(\displaystyle and \ du \ = \ \frac{xdx}{u}, \ \implies \ dx \ = \ \frac{udu}{x}\)
\(\displaystyle Hence, \ \int\frac{x^{3}u}{xu}du \ = \ \int x^{2}du \ = \ \int(u^{2}-1)du \ = \ \frac{u^{3}}{3}-u+C\)
\(\displaystyle Resub: \ \frac{(x^{2}+1)^{3/2}}{3}-(x^{2}+1)^{1/2}+C \ = \ \frac{(x^{2}+1)^{3/2}-3(x^{2}+1)^{1/2}}{3}+C\)
\(\displaystyle = \ \frac{(x^{2}+1)^{1/2}[(x^{2}+1)-3]}{3}+C \ = \ \frac{(x^{2}+1)^{1/2}(x^{2}-2)}{3}+C\)
\(\displaystyle Check: \ D_x\bigg[\frac{(x^{2}+1)^{1/2}(x^{2}-2)}{3}+C\bigg] \ = \ \frac{x^{3}}{(x^{2}+1)^{1/2}}, \ a \ good \ exercise \ for \ you.\)
Hello, serenity!
. . \(\displaystyle \begin{array}{ccccccc}u &=& x^2 && dv &=& x(x^2+1)^{-\frac{1}{2}}dx \\ \\[-3mm] du &=& 2x\,dx && v &=& (x^2+1)^{\frac{1}{2}} \end{array}\)
\(\displaystyle I \;\;=\;\;x^2(x^2+1)^{\frac{1}{2}} \:-\: \int 2x(x^2+1)^{\frac{1}{2}}\,dx\)
\(\displaystyle I \;\;=\;\;x^2(x^2+1)^{\frac{1}{2}} \:-\: \tfrac{2}{3}(x^2+1)^{\frac{3}{2}} + C\)
\(\displaystyle I \;=\;\int \frac{x^3}{\sqrt{x^2+1}}\,dx\)
. . \(\displaystyle \begin{array}{ccccccc}u &=& x^2 && dv &=& x(x^2+1)^{-\frac{1}{2}}dx \\ \\[-3mm] du &=& 2x\,dx && v &=& (x^2+1)^{\frac{1}{2}} \end{array}\)
\(\displaystyle I \;\;=\;\;x^2(x^2+1)^{\frac{1}{2}} \:-\: \int 2x(x^2+1)^{\frac{1}{2}}\,dx\)
\(\displaystyle I \;\;=\;\;x^2(x^2+1)^{\frac{1}{2}} \:-\: \tfrac{2}{3}(x^2+1)^{\frac{3}{2}} + C\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle serenity, \ for \ what \ it \ is \ worth \ dept.\)
\(\displaystyle If \ not \ I, \ follow \ the \ leads \ of \ the \ other \ moderators \ and \ forego \ tutorjoel \ (u \ = \ x^{3}, \ then \ du\)
\(\displaystyle \ = \ 3x \ dx?) \ as \ he \ is \ more \ interested \ in \ expounding \ on \ his \ erudition \ (which \ I \ think \ is \ sorrily\)
\(\displaystyle \ lacking) \ than \ aiding \ a \ future \ mathematician. \ In \ short, \ beware \ of \ people \ with \ appellations \ of\)
\(\displaystyle \ "doctor" \ or \ "tutor".\)
\(\displaystyle If \ not \ I, \ follow \ the \ leads \ of \ the \ other \ moderators \ and \ forego \ tutorjoel \ (u \ = \ x^{3}, \ then \ du\)
\(\displaystyle \ = \ 3x \ dx?) \ as \ he \ is \ more \ interested \ in \ expounding \ on \ his \ erudition \ (which \ I \ think \ is \ sorrily\)
\(\displaystyle \ lacking) \ than \ aiding \ a \ future \ mathematician. \ In \ short, \ beware \ of \ people \ with \ appellations \ of\)
\(\displaystyle \ "doctor" \ or \ "tutor".\)
tutor_joel
Junior Member
- Joined
- Feb 6, 2010
- Messages
- 157
That's the point of the forum. We all make mistakes. moderation. We are all learning.