What is the integral of xe^{x^2+1} ?

The integral of xe^{x^2+1} is 1/2 e^{x^2+1}+C

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integral of xe^{x^2+1}

\int x e^{x^{2} + 1} d x

\frac{1}{2} e^{x^{2} + 1} + C

Solution steps

\int x e^{x^{2} + 1} d x

Apply u-substitution

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

= \int \frac{e ^{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 e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{2} e^{u}

Substitute back

u = x^{2} + 1

= \frac{1}{2} e^{x^{2} + 1}

Add a constant to the solution

= \frac{1}{2} e^{x^{2} + 1} + C

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

d x + e^{4 x} d y = 0

\int \frac{8 ( x ^{2} + 1 )}{4 x ^{2} + 9} d x

\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )}

What is the integral of xe^{x^2+1} ?
The integral of xe^{x^2+1} is 1/2 e^{x^2+1}+C