Integration by parts: int (1/(x^(1/3)+x^(1/2))) dx
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BigGlenntheHeavy
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Re: Integration by parts
\(\displaystyle One \ way, \ a \ good \ workout \ with \ tex.\)
\(\displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} \ = \ \int\frac{dx}{x^{1/3}(1+x^{1/6})}\)
\(\displaystyle Let \ u \ = \ 1+x^{1/6}, \ then \ x^{1/6} \ = \ u-1 \ and \ du \ = \ \frac{dx}{6x^{5/6}} \ = \ \frac{dx}{6(u-1)^{5}}\)
\(\displaystyle So \ 6(u-1)^{5}du \ = \ dx \ and \ x^{1/3} \ = \ (u-1)^{2}.\)
\(\displaystyle Ergo, \ 6\int\frac{(u-1)^{5}du}{u(u-1)^{2}} \ = \ 6\int\frac{(u-1)^{3}du}{u} \ = \ 6\int\frac{(u^{3}-3u^{2}+3u-1)du}{u}\)
\(\displaystyle \ = \ 6\int(u^{2}-3u+3-\frac{1}{u})du \ = \ 6[\frac{u^{3}}{3}-\frac{3u^{2}}{2}+3u-ln|u|]+ C_{1}.\)
\(\displaystyle Substituting \ and \ expanding, \ we \ get \ x^{1/6}(2x^{1/3}-3x^{1/6}+6)-6ln(1+x^{1/6})+C, \ C \ = \ 11+C_{1}.\)
\(\displaystyle One \ way, \ a \ good \ workout \ with \ tex.\)
\(\displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} \ = \ \int\frac{dx}{x^{1/3}(1+x^{1/6})}\)
\(\displaystyle Let \ u \ = \ 1+x^{1/6}, \ then \ x^{1/6} \ = \ u-1 \ and \ du \ = \ \frac{dx}{6x^{5/6}} \ = \ \frac{dx}{6(u-1)^{5}}\)
\(\displaystyle So \ 6(u-1)^{5}du \ = \ dx \ and \ x^{1/3} \ = \ (u-1)^{2}.\)
\(\displaystyle Ergo, \ 6\int\frac{(u-1)^{5}du}{u(u-1)^{2}} \ = \ 6\int\frac{(u-1)^{3}du}{u} \ = \ 6\int\frac{(u^{3}-3u^{2}+3u-1)du}{u}\)
\(\displaystyle \ = \ 6\int(u^{2}-3u+3-\frac{1}{u})du \ = \ 6[\frac{u^{3}}{3}-\frac{3u^{2}}{2}+3u-ln|u|]+ C_{1}.\)
\(\displaystyle Substituting \ and \ expanding, \ we \ get \ x^{1/6}(2x^{1/3}-3x^{1/6}+6)-6ln(1+x^{1/6})+C, \ C \ = \ 11+C_{1}.\)
Re: Integration by parts
Hello, willc!
\(\displaystyle \text{Let: }\:x^{\frac{1}{6}} \:=\:u \quad\Rightarrow\quad x \:=\: u^6 \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:u^2 \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:u^3 \quad\Rightarrow\quad dx \:=\:6u^5\,du\)
\(\displaystyle \text{Substtute: }\;\int\frac{6u^5\,du}{u^2 + u^3} \;=\;6\int\frac{u^3}{1+u}\,du\)
\(\displaystyle \text{Long division: }\;6\int\left(u^2 - u + 1 - \frac{1}{u+1}\right)\,du\)
. . . . . . . . \(\displaystyle = \;6\left[\tfrac{1}{3}u^3 - \tfrac{1}{2}u^2 + u - \ln|u + 1|\right] + C\)
\(\displaystyle \text{Back-substitute: }\;6\bigg[\tfrac{1}{3}(x^{\frac{1}{6}})^3 - \tfrac{1}{2}(x^{\frac{1}{6}})^2 + x^{\frac{1}{6}} - \ln\left|x^{\frac{1}{6}} + 1\right| \bigg] + C\)
. . . . . . . . . . \(\displaystyle =\;6\bigg[\tfrac{1}{3}x^{\frac{1}{2}} - \tfrac{1}{2}x^{\frac{1}{3}} + x^{\frac{1}{6}} - \ln\left|x^{\frac{1}{6}} + 1\right|\bigg] + C\)
. . . . . . . . . . \(\displaystyle =\;2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\left|x^{\frac{1}{6}} + 1\right| + C\)
Hello, willc!
\(\displaystyle \text{Let: }\:x^{\frac{1}{6}} \:=\:u \quad\Rightarrow\quad x \:=\: u^6 \quad\Rightarrow\quad x^{\frac{1}{3}} \:=\:u^2 \quad\Rightarrow\quad x^{\frac{1}{2}} \:=\:u^3 \quad\Rightarrow\quad dx \:=\:6u^5\,du\)
\(\displaystyle \text{Substtute: }\;\int\frac{6u^5\,du}{u^2 + u^3} \;=\;6\int\frac{u^3}{1+u}\,du\)
\(\displaystyle \text{Long division: }\;6\int\left(u^2 - u + 1 - \frac{1}{u+1}\right)\,du\)
. . . . . . . . \(\displaystyle = \;6\left[\tfrac{1}{3}u^3 - \tfrac{1}{2}u^2 + u - \ln|u + 1|\right] + C\)
\(\displaystyle \text{Back-substitute: }\;6\bigg[\tfrac{1}{3}(x^{\frac{1}{6}})^3 - \tfrac{1}{2}(x^{\frac{1}{6}})^2 + x^{\frac{1}{6}} - \ln\left|x^{\frac{1}{6}} + 1\right| \bigg] + C\)
. . . . . . . . . . \(\displaystyle =\;6\bigg[\tfrac{1}{3}x^{\frac{1}{2}} - \tfrac{1}{2}x^{\frac{1}{3}} + x^{\frac{1}{6}} - \ln\left|x^{\frac{1}{6}} + 1\right|\bigg] + C\)
. . . . . . . . . . \(\displaystyle =\;2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\left|x^{\frac{1}{6}} + 1\right| + C\)