MHB What is the antiderivative of $\frac{1}{x^{1/2}+x^{1/3}}$ for $x>0$?

[lfdahl]

Find the antiderivative ($ x > 0$):\[\int\frac{1}{x^{1/2}+x^{1/3}}dx\]

 

[Dethrone]

Spoiler

If there are any mistakes -- apologies, doing this on my break.

Let $x=a^6 \implies dx=6a^5$
$$\int \frac{1}{x^{1/3}+x^{1/2}}dx=\int \frac{6a^5}{a^3+a^2}da=\int \frac{6a^3}{1+a}da$$
Let $b=a+1$
$$=\int \frac{6(b-1)^3}{b}db=6\int \frac{b^3-3b^2+3b-1}{b}db=6\int \left( b^2-3b+3-\frac{1}{b}\right) db$$
$$=6\left( \frac{b^3}{3}-\frac{3b^2}{2}+3b -\ln{b}\right)+C$$
$$=6\left( \frac{(1+a)^3}{3}-\frac{3(1+a)^2}{2}+3(1+a) -\ln{(1+a)}\right)+C$$
$$=6\left( \frac{(1+x^{1/6})^3}{3}-\frac{3(1+x^{1/6})^2}{2}+3(1+x^{1/6}) -\ln{(1+x^{1/6})}\right)+C$$

 

[SuperSonic4]

Find the antiderivative ($ x > 0$):\[\int\frac{1}{x^{1/2}+x^{1/3}}dx\]

Spoiler

Let $u = x^{1/6}$ hence $du = \dfrac{dx}{6x^{5/6}}$ and $dx = 6x^{5/6}du $

This can also be written in terms of $u$ as $dx = 6u^5 du$

Note that $x^{1/2} = x^{3/6} = u^3$ and also that $x^{1/3} = x^{2/6} = u^2$

Therefore the integral wrt u is now

$ 6\int \dfrac{u^5}{u^3 + u^2}$

I can cancel a $u^2$ as I am told that $x>0$ and so $u > 0$

$ 6 \int \dfrac{u^3}{u+1}$

Using long division to break this down (I don't know how to show this in Latex sorry)

$\dfrac{u^3}{u+1} = u^2-u+1-\dfrac{1}{u+1}$

Subsituting this back for the integrand gives

$6 \int \left(u^2 - u+1 - \dfrac{1}{u+1}\right) du = \int 6u^2 du - \int 6u du + \int 6du + \int \dfrac{6}{u+1} du$$2u^3 - 3u^2 + 6u + 6\ln (u+1) + C$ (I know that u is positive so no need for absolute value here)Back subbing for x

$2(x^{1/6})^3 - 3(x^{1/6})^2 + 6(x^{1/6}) + 6 \ln (x^{1/6} + 1) + C$

Tidy up a bit

$2x^{1/2} - 3x^{1/3} + 6x^{1/6} + 6\ln (x^{1/6} + 1) + C$

$2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} + 6 \ln (\sqrt[6]{x} + 1) + C$

 

[Prove It]

Find the antiderivative ($ x > 0$):\[\int\frac{1}{x^{1/2}+x^{1/3}}dx\]

Spoiler

$\displaystyle \begin{align*} \int{\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}\,\mathrm{d}x} &= \int{\frac{1}{x^{\frac{3}{6}} + x^{\frac{2}{6}}}\,\mathrm{d}x} \\ &= \int{ \frac{1}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\mathrm{d}x} \\ &= \int{ \frac{6\,x^{\frac{5}{6}}}{6\,x^{\frac{5}{6}}\,\left[ \left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2 \right] }\,\mathrm{d}x } \\ &= \int{\frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x} \end{align*}$

Now let $\displaystyle \begin{align*} u = x^{\frac{1}{6}} \implies \mathrm{d}u = \frac{1}{6\,x^{\frac{5}{6}}}\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x } &= \int{\frac{6\,u^5}{u^3 + u^2}\,\mathrm{d}u} \\ &= \int{ \frac{6\,u^3}{u + 1}\,\mathrm{d}u } \\ &= 6 \int{ \left( u^2 - u + 1 - \frac{1}{u + 1}\right) \,\mathrm{d}u } \\ &= 6\,\left( \frac{u^3}{3} - \frac{u^2}{2} + u - \ln{ \left| u + 1 \right| } \right) + C \\ &= 2\,u^3 - 3\,u^2 + 6\,u - 6\ln{ \left| u + 1 \right| } + C \\ &= 2\,\left( x^{\frac{1}{6}} \right) ^3 - 3\,\left( x^{\frac{1}{6}} \right) ^2 + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \\ &= 2\,x^{\frac{1}{2}} - 3\,x^{\frac{1}{3}} + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \end{align*}$

 

[lfdahl]

Spoiler

If there are any mistakes -- apologies, doing this on my break.

Let $x=a^6 \implies dx=6a^5$
$$\int \frac{1}{x^{1/3}+x^{1/2}}dx=\int \frac{6a^5}{a^3+a^2}da=\int \frac{6a^3}{1+a}da$$
Let $b=a+1$
$$=\int \frac{6(b-1)^3}{b}db=6\int \frac{b^3-3b^2+3b-1}{b}db=6\int \left( b^2-3b+3-\frac{1}{b}\right) db$$
$$=6\left( \frac{b^3}{3}-\frac{3b^2}{2}+3b -\ln{b}\right)+C$$
$$=6\left( \frac{(1+a)^3}{3}-\frac{3(1+a)^2}{2}+3(1+a) -\ln{(1+a)}\right)+C$$
$$=6\left( \frac{(1+x^{1/6})^3}{3}-\frac{3(1+x^{1/6})^2}{2}+3(1+x^{1/6}) -\ln{(1+x^{1/6})}\right)+C$$

Hi, Rido12

Thankyou for your solution, your method is correct:

Spoiler

- but you forgot to reduce the expression: \[ \frac{(1+a)^3}{3}-\frac{3(1+a)^2}{2}+3(1+a)\]

- - - Updated - - -

Spoiler

Let $u = x^{1/6}$ hence $du = \dfrac{dx}{6x^{5/6}}$ and $dx = 6x^{5/6}du $

This can also be written in terms of $u$ as $dx = 6u^5 du$

Note that $x^{1/2} = x^{3/6} = u^3$ and also that $x^{1/3} = x^{2/6} = u^2$

Therefore the integral wrt u is now

$ 6\int \dfrac{u^5}{u^3 + u^2}$

I can cancel a $u^2$ as I am told that $x>0$ and so $u > 0$

$ 6 \int \dfrac{u^3}{u+1}$

Using long division to break this down (I don't know how to show this in Latex sorry)

$\dfrac{u^3}{u+1} = u^2-u+1-\dfrac{1}{u+1}$

Subsituting this back for the integrand gives

$6 \int \left(u^2 - u+1 - \dfrac{1}{u+1}\right) du = \int 6u^2 du - \int 6u du + \int 6du + \int \dfrac{6}{u+1} du$$2u^3 - 3u^2 + 6u + 6\ln (u+1) + C$ (I know that u is positive so no need for absolute value here)Back subbing for x

$2(x^{1/6})^3 - 3(x^{1/6})^2 + 6(x^{1/6}) + 6 \ln (x^{1/6} + 1) + C$

Tidy up a bit

$2x^{1/2} - 3x^{1/3} + 6x^{1/6} + 6\ln (x^{1/6} + 1) + C$

$2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} + 6 \ln (\sqrt[6]{x} + 1) + C$

Good job, SuperSonic4! Thankyou for your participation!

- - - Updated - - -

Spoiler

$\displaystyle \begin{align*} \int{\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}\,\mathrm{d}x} &= \int{\frac{1}{x^{\frac{3}{6}} + x^{\frac{2}{6}}}\,\mathrm{d}x} \\ &= \int{ \frac{1}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\mathrm{d}x} \\ &= \int{ \frac{6\,x^{\frac{5}{6}}}{6\,x^{\frac{5}{6}}\,\left[ \left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2 \right] }\,\mathrm{d}x } \\ &= \int{\frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x} \end{align*}$

Now let $\displaystyle \begin{align*} u = x^{\frac{1}{6}} \implies \mathrm{d}u = \frac{1}{6\,x^{\frac{5}{6}}}\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x } &= \int{\frac{6\,u^5}{u^3 + u^2}\,\mathrm{d}u} \\ &= \int{ \frac{6\,u^3}{u + 1}\,\mathrm{d}u } \\ &= 6 \int{ \left( u^2 - u + 1 - \frac{1}{u + 1}\right) \,\mathrm{d}u } \\ &= 6\,\left( \frac{u^3}{3} - \frac{u^2}{2} + u - \ln{ \left| u + 1 \right| } \right) + C \\ &= 2\,u^3 - 3\,u^2 + 6\,u - 6\ln{ \left| u + 1 \right| } + C \\ &= 2\,\left( x^{\frac{1}{6}} \right) ^3 - 3\,\left( x^{\frac{1}{6}} \right) ^2 + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \\ &= 2\,x^{\frac{1}{2}} - 3\,x^{\frac{1}{3}} + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \end{align*}$

Awesome, Prove It! Thankyou for your participation!