# How do you find the indefinite integral of (10e^(-2x^5))/x^6 dx?

Let $f : \mathbb{R} \setminus \left\{0\right\} \rightarrow \mathbb{R}$ such that $f \left(x\right) : = \frac{10 {e}^{- 2 {x}^{5}}}{{x}^{6}}$. We want to calculate the antiderivative $F$ of the function $f$, so
$F \left(x\right) : = \int f \left(x\right) \mathrm{dx}$.
In answering to the specific problem, I'll try to illustrate a way one can approach a problem like this.

The two main (elementary) integration techniques are integration by substitution and integration by parts .

Integration by substitution is often used in presence of composition of functions (functions "nested" in other functions) and we want to try to get rid of this "nested" structure. This can be the case. In particular, the exponent of ${e}^{- 2 {x}^{5}}$ can be seen as the function $t \left(x\right) : = - 2 {x}^{5}$ and so the exponential becomes ${e}^{t \left(x\right)}$. The function we chose is invertible, and its inverse is $x \left(t\right) : = - \sqrt[5]{\frac{1}{2} t}$. This inverse allows us to change variable from $x$ to $t$:
$f \left(x \left(t\right)\right) = \frac{10 {e}^{- 2 {\left[x \left(t\right)\right]}^{5}}}{{\left[x \left(t\right)\right]}^{6}} = \frac{10 {e}^{t}}{{\left(- \sqrt[5]{\frac{1}{2} t}\right)}^{6}} = \frac{10 {e}^{t}}{{\left(\frac{1}{2} t\right)}^{\frac{6}{5}}}$
And to rewrite the differential $\mathrm{dx}$ in terms of the differential $\mathrm{dt}$
$\mathrm{dx} = \frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} \mathrm{dt} = - \frac{1}{2} {\left(\frac{1}{2} t\right)}^{- \frac{4}{5}} \mathrm{dt}$
So we get
$F \left(x \left(t\right)\right) : = - \frac{1}{2} \int \frac{10 {e}^{t}}{{\left(\frac{1}{2} t\right)}^{\frac{6}{5}}} {\left(\frac{1}{2} t\right)}^{- \frac{4}{5}} \mathrm{dt} = - 5 \int \frac{{e}^{t}}{{\left(\frac{1}{2} t\right)}^{2}} \mathrm{dt} = - 20 \int \frac{{e}^{t}}{{t}^{2}} \mathrm{dt}$
The integral of the last function cannot be expressed in terms of a finite representation of elementary functions. This means that $F \left(x\right)$ can't be expressed using a finite number of elemetary functions, otherwise we could obtain $F \left(x\right)$ from $F \left(x \left(t\right)\right)$ by substituting $t = t \left(x\right)$.

Although we already got an answer, let's try by parts. Integration by parts comes from the derivative of a product of functions:
$\frac{d}{\mathrm{dx}} \left[a \left(x\right) b \left(x\right)\right] = \frac{d}{\mathrm{dx}} \left[a \left(x\right)\right] b \left(x\right) + a \left(x\right) \frac{d}{\mathrm{dx}} \left[b \left(x\right)\right]$
So we have to see $f$ as a product of some derivative $a ' \left(x\right)$ and some function $b ' \left(x\right)$ to be derived. A natural choice would be
$a ' \left(x\right) : = \frac{1}{{x}^{6}} = {x}^{- 6}$
$b \left(x\right) : = 10 {e}^{- 2 {x}^{5}}$
because it's pretty easy to compute the antiderivative of $a ' \left(x\right)$.
$a \left(x\right) = \frac{{x}^{- 5}}{- 5} = - \frac{1}{5 {x}^{5}}$
$b ' \left(x\right) = 10 \left(- 10 {x}^{4}\right) {e}^{- 2 {x}^{5}} = - 100 {x}^{4} {e}^{- 2 {x}^{5}}$
We get
$F \left(x\right) : = \int a ' \left(x\right) b \left(x\right) \mathrm{dx} = a \left(x\right) b \left(x\right) - \int a \left(x\right) b ' \left(x\right) \mathrm{dx} = - \frac{2 {e}^{- 2 {x}^{5}}}{{x}^{5}} - 20 \int \frac{{e}^{- 2 {x}^{5}}}{x} \mathrm{dx}$
We could integrate this integral by parts again, considering
$t i l \mathrm{de} \left\{a\right\} ' \left(x\right) = \frac{1}{x}$
$b \left(x\right) = {e}^{- 2 {x}^{5}}$
but this would introduce logarithms and won't simplify the calculations.

One can express the solution by integral functions or try to write a series expansion of it.