What is the integral of x0.2^{x^2} ?

The integral of x0.2^{x^2} is -0.31066…*0.2^{x^2}+C

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integral of x0.2^{x^2}

\int x 0. 2^{x^{2}} d x

- 0.31066 \dots \cdot 0. 2^{x^{2}} + C

Solution steps

\int x \cdot 0. 2^{x^{2}} d x

Apply u-substitution

: \int \frac{0. 2 ^{u}}{2} d u

= \int \frac{0. 2 ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 0. 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{2} \cdot \frac{0. 2 ^{u}}{ln ( 0.2 )}

Substitute back

u = x^{2}

= \frac{1}{2} \cdot \frac{0. 2 ^{x^{2}}}{ln ( 0.2 )}

Simplify

\frac{1}{2} \cdot \frac{0. 2 ^{x^{2}}}{ln ( 0.2 )} : - 0.31066 \dots \cdot 0. 2^{x^{2}}

= - 0.31066 \dots \cdot 0. 2^{x^{2}}

Add a constant to the solution

= - 0.31066 \dots \cdot 0. 2^{x^{2}} + C

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What is the integral of x0.2^{x^2} ?
The integral of x0.2^{x^2} is -0.31066…*0.2^{x^2}+C