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

The integral of (x+1)4^{(x+1)^2} is 8(x+1)^2+C

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

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

8 (x + 1)^{2} + C

Solution steps

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

Apply u-substitution

: \int 8 d u

= \int 8 d u

Integral of a constant:

\int a d x = a x

= 8 u

Substitute back

u = (x + 1)^{2}

= 8 (x + 1)^{2}

Add a constant to the solution

= 8 (x + 1)^{2} + C

\int x^{3} ln (4 x^{2}) d x

\int \frac{1}{ln ( x - 1 )} d x

\int \frac{1}{x ^{2} 2 - 3 x} d x

\int (\frac{1 + ln ( x )}{x}) d x

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

What is the integral of (x+1)4^{(x+1)^2} ?
The integral of (x+1)4^{(x+1)^2} is 8(x+1)^2+C