What is the integral of 10^{x^2}x ?

The integral of 10^{x^2}x is (5^{x^2}*2^{x^2-1})/(ln(10))+C

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integral of 10^{x^2}x

\int 1 0^{x^{2}} x d x

\frac{5 ^{x^{2}} \cdot 2 ^{x^{2} - 1}}{ln ( 10 )} + C

Solution steps

\int 1 0^{x^{2}} x d x

Apply u-substitution

: \int 5^{u} \cdot 2^{u - 1} d u

= \int 5^{u} \cdot 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= \int 5^{u} \cdot 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= 2^{- 1} \cdot \int 5^{u} \cdot 2^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{2} \cdot \int 5^{u} \cdot 2^{u} d u

Simplify

5^{u} \cdot 2^{u} : 1 0^{u}

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

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

= \frac{1}{2} \cdot \frac{1 0 ^{u}}{ln ( 10 )}

Substitute back

u = x^{2}

= \frac{1}{2} \cdot \frac{1 0 ^{x^{2}}}{ln ( 10 )}

Simplify

\frac{1}{2} \cdot \frac{1 0 ^{x^{2}}}{ln ( 10 )} : \frac{5 ^{x^{2}} \cdot 2 ^{x^{2} - 1}}{ln ( 10 )}

= \frac{5 ^{x^{2}} \cdot 2 ^{x^{2} - 1}}{ln ( 10 )}

Add a constant to the solution

= \frac{5 ^{x^{2}} \cdot 2 ^{x^{2} - 1}}{ln ( 10 )} + C

x \to - \infty lim (x^{3} e^{x})

x \to - 2 - lim (f (x) x)

x \to 0 lim (\frac{2}{x ^{3} + x})

y = π

\frac{\partial}{\partial x} (x^{3} y^{4} + x^{4} y^{3})

What is the integral of 10^{x^2}x ?
The integral of 10^{x^2}x is (5^{x^2}*2^{x^2-1})/(ln(10))+C