What is the integral of (x+4)6^{(x+4)^2} ?

The integral of (x+4)6^{(x+4)^2} is (3^{(x+4)^2}*2^{x^2+8x+15})/(ln(6))+C

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integral of (x+4)6^{(x+4)^2}

\int (x + 4) 6^{(x + 4)^{2}} d x

\frac{3 ^{(x + 4)^{2}} \cdot 2 ^{x^{2} + 8 x + 15}}{ln ( 6 )} + C

Solution steps

\int (x + 4) \cdot 6^{(x + 4)^{2}} d x

Apply u-substitution

: \int 6^{u^{2}} u d u

= \int 6^{u^{2}} u d u

Apply u-substitution

: \int 3^{v} \cdot 2^{v - 1} d v

= \int 3^{v} \cdot 2^{v - 1} d v

Simplify

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

= \int 3^{v} \cdot 2^{v} \cdot 2^{- 1} d v

Take the constant out:

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

= 2^{- 1} \cdot \int 3^{v} \cdot 2^{v} d v

Apply exponent rule:

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

= \frac{1}{2} \cdot \int 3^{v} \cdot 2^{v} d v

Simplify

3^{v} \cdot 2^{v} : 6^{v}

= \frac{1}{2} \cdot \int 6^{v} d v

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

= \frac{1}{2} \cdot \frac{6 ^{v}}{ln ( 6 )}

Substitute back

= \frac{1}{2} \cdot \frac{6 ^{(x + 4)^{2}}}{ln ( 6 )}

Simplify

\frac{1}{2} \cdot \frac{6 ^{(x + 4)^{2}}}{ln ( 6 )} : \frac{3 ^{(x + 4)^{2}} \cdot 2 ^{x^{2} + 8 x + 15}}{ln ( 6 )}

= \frac{3 ^{(x + 4)^{2}} \cdot 2 ^{x^{2} + 8 x + 15}}{ln ( 6 )}

Add a constant to the solution

= \frac{3 ^{(x + 4)^{2}} \cdot 2 ^{x^{2} + 8 x + 15}}{ln ( 6 )} + C

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What is the integral of (x+4)6^{(x+4)^2} ?
The integral of (x+4)6^{(x+4)^2} is (3^{(x+4)^2}*2^{x^2+8x+15})/(ln(6))+C