What is the integral of x^{4035}e^{x^{2018}} ?

The integral of x^{4035}e^{x^{2018}} is 1/2018 (e^{x^{2018}}x^{2018}-e^{x^{2018}})+C

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integral of x^{4035}e^{x^{2018}}

\int x^{4035} e^{x^{2018}} d x

\frac{1}{2018} (e^{x^{2018}} x^{2018} - e^{x^{2018}}) + C

Solution steps

\int x^{4035} e^{x^{2018}} d x

Apply u-substitution

: \int \frac{ln ( u )}{2018} d u

= \int \frac{ln ( u )}{2018} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2018} \cdot \int ln (u) d u

Apply Integration By Parts

: u ln (u) - \int 1 d u

= \frac{1}{2018} (u ln (u) - \int 1 d u)

\int 1 d u = u

= \frac{1}{2018} (u ln (u) - u)

Substitute back

u = e^{x^{2018}}

= \frac{1}{2018} (e^{x^{2018}} ln (e^{x^{2018}}) - e^{x^{2018}})

Simplify

ln (e^{x^{2018}}) : x^{2018}

= \frac{1}{2018} (e^{x^{2018}} x^{2018} - e^{x^{2018}})

Add a constant to the solution

= \frac{1}{2018} (e^{x^{2018}} x^{2018} - e^{x^{2018}}) + C

x \to 1 lim (x arcsin (\frac{x}{2}))

\frac{d}{d x} (3 \cdot 1 0^{2 x})

\int t ln (t + 1) d t

\frac{\partial}{\partial x} (2 (1 + y) x)

y = 3^{x^{2} + 3 x}

What is the integral of x^{4035}e^{x^{2018}} ?
The integral of x^{4035}e^{x^{2018}} is 1/2018 (e^{x^{2018}}x^{2018}-e^{x^{2018}})+C