What is the integral of (x+1)6^{(x+1)^2} ?

The integral of (x+1)6^{(x+1)^2} is (3^{(x+1)^2}*2^{x^2+2x})/(ln(6))+C

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integral of (x+1)6^{(x+1)^2}

\int (x + 1) 6^{(x + 1)^{2}} d x

\frac{3 ^{(x + 1)^{2}} \cdot 2 ^{x^{2} + 2 x}}{ln ( 6 )} + C

Solution steps

\int (x + 1) \cdot 6^{(x + 1)^{2}} d x

Apply u-substitution

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

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

Simplify

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

= \int 3^{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 3^{u} \cdot 2^{u} d u

Apply exponent rule:

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

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

Simplify

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

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

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

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

Substitute back

u = (x + 1)^{2}

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

Simplify

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

= \frac{3 ^{(x + 1)^{2}} \cdot 2 ^{x^{2} + 2 x}}{ln ( 6 )}

Add a constant to the solution

= \frac{3 ^{(x + 1)^{2}} \cdot 2 ^{x^{2} + 2 x}}{ln ( 6 )} + C

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What is the integral of (x+1)6^{(x+1)^2} ?
The integral of (x+1)6^{(x+1)^2} is (3^{(x+1)^2}*2^{x^2+2x})/(ln(6))+C