What is the integral of 5^{x^2+2x+3}(x+1) ?

The integral of 5^{x^2+2x+3}(x+1) is (5^{x^2+2x+3})/(2ln(5))+C

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integral of 5^{x^2+2x+3}(x+1)

\int 5^{x^{2} + 2 x + 3} (x + 1) d x

\frac{5 ^{x^{2} + 2 x + 3}}{2 ln ( 5 )} + C

Solution steps

\int 5^{x^{2} + 2 x + 3} (x + 1) d x

Apply u-substitution

: \int 5^{u^{2} + 2} u d u

= \int 5^{u^{2} + 2} u d u

Apply u-substitution

: \int \frac{5 ^{v}}{2} d v

= \int \frac{5 ^{v}}{2} d v

Take the constant out:

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

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

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

= \frac{1}{2} \cdot \frac{5 ^{v}}{ln ( 5 )}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{5 ^{(x + 1)^{2} + 2}}{ln ( 5 )} : \frac{5 ^{x^{2} + 2 x + 3}}{2 ln ( 5 )}

= \frac{5 ^{x^{2} + 2 x + 3}}{2 ln ( 5 )}

Add a constant to the solution

= \frac{5 ^{x^{2} + 2 x + 3}}{2 ln ( 5 )} + C

\int 5 x^{\frac{1}{4}} d x

\frac{s}{s ^{2} + \frac{1}{l c}}

x \to 2 lim (x^{2} f (x))

\int_{0}^{1} x \cdot sin (π x) d x

y = sec (3 x^{4} + 5)

What is the integral of 5^{x^2+2x+3}(x+1) ?
The integral of 5^{x^2+2x+3}(x+1) is (5^{x^2+2x+3})/(2ln(5))+C