What is the integral of 5^{4x^2+4x+1}(2x+1) ?

The integral of 5^{4x^2+4x+1}(2x+1) is 1/(4ln(5))*5^{(2x+1)^2}+C

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integral of 5^{4x^2+4x+1}(2x+1)

\int 5^{4 x^{2} + 4 x + 1} (2 x + 1) d x

\frac{1}{4 ln ( 5 )} \cdot 5^{(2 x + 1)^{2}} + C

Solution steps

\int 5^{4 x^{2} + 4 x + 1} (2 x + 1) d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{2 ln ( 5 )} d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2 ln ( 5 )} \cdot 5^{(2 x + 1)^{2}} : \frac{1}{4 ln ( 5 )} \cdot 5^{(2 x + 1)^{2}}

= \frac{1}{4 ln ( 5 )} \cdot 5^{(2 x + 1)^{2}}

Add a constant to the solution

= \frac{1}{4 ln ( 5 )} \cdot 5^{(2 x + 1)^{2}} + C

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What is the integral of 5^{4x^2+4x+1}(2x+1) ?
The integral of 5^{4x^2+4x+1}(2x+1) is 1/(4ln(5))*5^{(2x+1)^2}+C