What is the integral of e^{(1+n)x} ?

The integral of e^{(1+n)x} is e^{(1+n)x}x-(e^{x(n+1)}x(n+1)-e^{x(n+1)})/(1+n)+C

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integral of e^{(1+n)x}

\int e^{(1 + n) x} d x

e^{(1 + n) x} x - \frac{e ^{x (n + 1)} x ( n + 1 ) - e ^{x (n + 1)}}{1 + n} + C

Solution steps

\int e^{(1 + n) x} d x

Apply Integration By Parts

: e^{(1 + n) x} x - \int e^{(1 + n) x} (1 + n) x d x

= e^{(1 + n) x} x - \int e^{(1 + n) x} (1 + n) x d x

\int e^{(1 + n) x} (1 + n) x d x = \frac{e ^{x (n + 1)} x ( n + 1 ) - e ^{x (n + 1)}}{1 + n}

= e^{(1 + n) x} x - \frac{e ^{x (n + 1)} x ( n + 1 ) - e ^{x (n + 1)}}{1 + n}

Add a constant to the solution

= e^{(1 + n) x} x - \frac{e ^{x (n + 1)} x ( n + 1 ) - e ^{x (n + 1)}}{1 + n} + C

\int \frac{x}{x + 1} d x

f (x) = 3 x + 1, at x = 8

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

p (t) = 25 e^{0.03 t}

\int \frac{( x ^{3} - 1 )}{( 4 x ^{3} - x )} d x

What is the integral of e^{(1+n)x} ?
The integral of e^{(1+n)x} is e^{(1+n)x}x-(e^{x(n+1)}x(n+1)-e^{x(n+1)})/(1+n)+C