What is the integral of s10^{s^2} ?

The integral of s10^{s^2} is (5^{s^2}*2^{s^2-1})/(ln(10))+C

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integral of s10^{s^2}

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

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

Solution steps

\int s \cdot 1 0^{s^{2}} d s

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 = s^{2}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of s10^{s^2} ?
The integral of s10^{s^2} is (5^{s^2}*2^{s^2-1})/(ln(10))+C