Integrate (1/(x^(1/3)+x^(1/2)))dx
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Jun 3, 2009 #2 Re: Integration by parts Why do think the antiderivative exists in a simple form? This may lead to something. x1/3+x1/2=(x1/6)2⋅(x1/6+1)\displaystyle x^{1/3} + x^{1/2} = \left( x^{1/6} \right)^{2} \cdot \left( x^{1/6}+1 \right)x1/3+x1/2=(x1/6)2⋅(x1/6+1) Or, maybe it won't.
Re: Integration by parts Why do think the antiderivative exists in a simple form? This may lead to something. x1/3+x1/2=(x1/6)2⋅(x1/6+1)\displaystyle x^{1/3} + x^{1/2} = \left( x^{1/6} \right)^{2} \cdot \left( x^{1/6}+1 \right)x1/3+x1/2=(x1/6)2⋅(x1/6+1) Or, maybe it won't.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jun 3, 2009 #3 Re: Integration by parts One way, a good workout with tex.\displaystyle One \ way, \ a \ good \ workout \ with \ tex.One way, a good workout with tex. ∫dxx1/2+x1/3 = ∫dxx1/3(1+x1/6)\displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} \ = \ \int\frac{dx}{x^{1/3}(1+x^{1/6})}∫x1/2+x1/3dx = ∫x1/3(1+x1/6)dx Let u = 1+x1/6, then x1/6 = u−1 and du = dx6x5/6 = dx6(u−1)5\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}}Let u = 1+x1/6, then x1/6 = u−1 and du = 6x5/6dx = 6(u−1)5dx So 6(u−1)5du = dx and x1/3 = (u−1)2.\displaystyle So \ 6(u-1)^{5}du \ = \ dx \ and \ x^{1/3} \ = \ (u-1)^{2}.So 6(u−1)5du = dx and x1/3 = (u−1)2. Ergo, 6∫(u−1)5duu(u−1)2 = 6∫(u−1)3duu = 6∫(u3−3u2+3u−1)duu\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}Ergo, 6∫u(u−1)2(u−1)5du = 6∫u(u−1)3du = 6∫u(u3−3u2+3u−1)du = 6∫(u2−3u+3−1u)du = 6[u33−3u22+3u−ln∣u∣]+C1.\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}. = 6∫(u2−3u+3−u1)du = 6[3u3−23u2+3u−ln∣u∣]+C1. Substituting and expanding, we get x1/6(2x1/3−3x1/6+6)−6ln(1+x1/6)+C, C = 11+C1.\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}.Substituting and expanding, we get x1/6(2x1/3−3x1/6+6)−6ln(1+x1/6)+C, C = 11+C1.
Re: Integration by parts One way, a good workout with tex.\displaystyle One \ way, \ a \ good \ workout \ with \ tex.One way, a good workout with tex. ∫dxx1/2+x1/3 = ∫dxx1/3(1+x1/6)\displaystyle \int\frac{dx}{x^{1/2}+x^{1/3}} \ = \ \int\frac{dx}{x^{1/3}(1+x^{1/6})}∫x1/2+x1/3dx = ∫x1/3(1+x1/6)dx Let u = 1+x1/6, then x1/6 = u−1 and du = dx6x5/6 = dx6(u−1)5\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}}Let u = 1+x1/6, then x1/6 = u−1 and du = 6x5/6dx = 6(u−1)5dx So 6(u−1)5du = dx and x1/3 = (u−1)2.\displaystyle So \ 6(u-1)^{5}du \ = \ dx \ and \ x^{1/3} \ = \ (u-1)^{2}.So 6(u−1)5du = dx and x1/3 = (u−1)2. Ergo, 6∫(u−1)5duu(u−1)2 = 6∫(u−1)3duu = 6∫(u3−3u2+3u−1)duu\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}Ergo, 6∫u(u−1)2(u−1)5du = 6∫u(u−1)3du = 6∫u(u3−3u2+3u−1)du = 6∫(u2−3u+3−1u)du = 6[u33−3u22+3u−ln∣u∣]+C1.\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}. = 6∫(u2−3u+3−u1)du = 6[3u3−23u2+3u−ln∣u∣]+C1. Substituting and expanding, we get x1/6(2x1/3−3x1/6+6)−6ln(1+x1/6)+C, C = 11+C1.\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}.Substituting and expanding, we get x1/6(2x1/3−3x1/6+6)−6ln(1+x1/6)+C, C = 11+C1.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jun 3, 2009 #4 Re: Integration by parts Hello, willc! Integrate: ∫dxx13+x12\displaystyle \text{Integrate: }\;\int\frac{dx}{x^{\frac{1}{3}} + x^{\frac{1}{2}}}Integrate: ∫x31+x21dx Click to expand... Let: x16 = u⇒x = u6⇒x13 = u2⇒x12 = u3⇒dx = 6u5 du\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\,duLet: x61=u⇒x=u6⇒x31=u2⇒x21=u3⇒dx=6u5du Substtute: ∫6u5 duu2+u3 = 6∫u31+u du\displaystyle \text{Substtute: }\;\int\frac{6u^5\,du}{u^2 + u^3} \;=\;6\int\frac{u^3}{1+u}\,duSubsttute: ∫u2+u36u5du=6∫1+uu3du Long division: 6∫(u2−u+1−1u+1) du\displaystyle \text{Long division: }\;6\int\left(u^2 - u + 1 - \frac{1}{u+1}\right)\,duLong division: 6∫(u2−u+1−u+11)du . . . . . . . . = 6[13u3−12u2+u−ln∣u+1∣]+C\displaystyle = \;6\left[\tfrac{1}{3}u^3 - \tfrac{1}{2}u^2 + u - \ln|u + 1|\right] + C=6[31u3−21u2+u−ln∣u+1∣]+C Back-substitute: 6[13(x16)3−12(x16)2+x16−ln∣x16+1∣]+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] + CBack-substitute: 6[31(x61)3−21(x61)2+x61−ln∣∣∣∣x61+1∣∣∣∣]+C . . . . . . . . . . = 6[13x12−12x13+x16−ln∣x16+1∣]+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=6[31x21−21x31+x61−ln∣∣∣∣x61+1∣∣∣∣]+C . . . . . . . . . . = 2x12−3x13+6x16−6∣x16+1∣+C\displaystyle =\;2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\left|x^{\frac{1}{6}} + 1\right| + C=2x21−3x31+6x61−6∣∣∣∣x61+1∣∣∣∣+C
Re: Integration by parts Hello, willc! Integrate: ∫dxx13+x12\displaystyle \text{Integrate: }\;\int\frac{dx}{x^{\frac{1}{3}} + x^{\frac{1}{2}}}Integrate: ∫x31+x21dx Click to expand... Let: x16 = u⇒x = u6⇒x13 = u2⇒x12 = u3⇒dx = 6u5 du\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\,duLet: x61=u⇒x=u6⇒x31=u2⇒x21=u3⇒dx=6u5du Substtute: ∫6u5 duu2+u3 = 6∫u31+u du\displaystyle \text{Substtute: }\;\int\frac{6u^5\,du}{u^2 + u^3} \;=\;6\int\frac{u^3}{1+u}\,duSubsttute: ∫u2+u36u5du=6∫1+uu3du Long division: 6∫(u2−u+1−1u+1) du\displaystyle \text{Long division: }\;6\int\left(u^2 - u + 1 - \frac{1}{u+1}\right)\,duLong division: 6∫(u2−u+1−u+11)du . . . . . . . . = 6[13u3−12u2+u−ln∣u+1∣]+C\displaystyle = \;6\left[\tfrac{1}{3}u^3 - \tfrac{1}{2}u^2 + u - \ln|u + 1|\right] + C=6[31u3−21u2+u−ln∣u+1∣]+C Back-substitute: 6[13(x16)3−12(x16)2+x16−ln∣x16+1∣]+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] + CBack-substitute: 6[31(x61)3−21(x61)2+x61−ln∣∣∣∣x61+1∣∣∣∣]+C . . . . . . . . . . = 6[13x12−12x13+x16−ln∣x16+1∣]+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=6[31x21−21x31+x61−ln∣∣∣∣x61+1∣∣∣∣]+C . . . . . . . . . . = 2x12−3x13+6x16−6∣x16+1∣+C\displaystyle =\;2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6\left|x^{\frac{1}{6}} + 1\right| + C=2x21−3x31+6x61−6∣∣∣∣x61+1∣∣∣∣+C
