lfdahl said:Find the antiderivative ($ x > 0$):\[\int\frac{1}{x^{1/2}+x^{1/3}}dx\]
lfdahl said:Find the antiderivative ($ x > 0$):\[\int\frac{1}{x^{1/2}+x^{1/3}}dx\]
Rido12 said: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 said: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
$2x^{1/2} - 3x^{1/3} + 6x^{1/6} + 6\ln (x^{1/6} + 1) + C$
$2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} + 6 \ln (\sqrt[6]{x} + 1) + C$
Awesome, Prove It! Thankyou for your participation!Prove It said:$\displaystyle \begin{align*} \int{\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}\,\mathrm{d}x} &= \int{\frac{1}{x^{\frac{3}{6}} + x^{\frac{2}{6}}}\,\mathrm{d}x} \\ &= \int{ \frac{1}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\mathrm{d}x} \\ &= \int{ \frac{6\,x^{\frac{5}{6}}}{6\,x^{\frac{5}{6}}\,\left[ \left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2 \right] }\,\mathrm{d}x } \\ &= \int{\frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x} \end{align*}$
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
$\displaystyle \begin{align*} \int{ \frac{6\,\left( x^{\frac{1}{6}} \right) ^5}{\left( x^{\frac{1}{6}} \right) ^3 + \left( x^{\frac{1}{6}} \right) ^2}\,\left( \frac{1}{6\,x^{\frac{5}{6}}} \right) \,\mathrm{d}x } &= \int{\frac{6\,u^5}{u^3 + u^2}\,\mathrm{d}u} \\ &= \int{ \frac{6\,u^3}{u + 1}\,\mathrm{d}u } \\ &= 6 \int{ \left( u^2 - u + 1 - \frac{1}{u + 1}\right) \,\mathrm{d}u } \\ &= 6\,\left( \frac{u^3}{3} - \frac{u^2}{2} + u - \ln{ \left| u + 1 \right| } \right) + C \\ &= 2\,u^3 - 3\,u^2 + 6\,u - 6\ln{ \left| u + 1 \right| } + C \\ &= 2\,\left( x^{\frac{1}{6}} \right) ^3 - 3\,\left( x^{\frac{1}{6}} \right) ^2 + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \\ &= 2\,x^{\frac{1}{2}} - 3\,x^{\frac{1}{3}} + 6\,x^{\frac{1}{6}} - 6\ln{ \left| x^{\frac{1}{6}} + 1 \right| } + C \end{align*}$
An antiderivative is the inverse operation of a derivative. It is a function that, when differentiated, yields the original function. In other words, it is the function that "undoes" the process of differentiation.
Finding the antiderivative is important because it allows us to solve problems involving rates of change, such as finding the distance traveled by an object given its velocity. It also helps us to understand the behavior of a function and make predictions about its values.
To find the antiderivative of a function, you can use techniques such as integration by substitution or integration by parts. It is also helpful to memorize common antiderivatives and their corresponding derivatives.
No, not every function can be integrated to find its antiderivative. Some functions, such as irrational or transcendental functions, do not have an antiderivative that can be expressed in terms of elementary functions. In these cases, we can use numerical methods or approximate the antiderivative.
You can check if you have found the correct antiderivative by differentiating the function and seeing if it matches the original function. Additionally, you can use the fundamental theorem of calculus to evaluate the definite integral of the function and compare it to the result of finding the antiderivative.