What is the integral of 3sqrt(x)e^{sqrt(x)} ?

The integral of 3sqrt(x)e^{sqrt(x)} is 6(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C

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integral of 3sqrt(x)e^{sqrt(x)}

\int 3 x e^{x} d x

6 (e^{x} x - 2 (e^{x} x - e^{x})) + C

Solution steps

\int 3 x e^{x} d x

Take the constant out:

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

= 3 \cdot \int x e^{x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

= 3 \cdot 2 (u^{2} e^{u} - \int 2 u e^{u} d u)

\int 2 u e^{u} d u = 2 (e^{u} u - e^{u})

= 3 \cdot 2 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = x

= 3 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x}))

Simplify

3 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x})) : 6 (e^{x} x - 2 (e^{x} x - e^{x}))

= 6 (e^{x} x - 2 (e^{x} x - e^{x}))

Add a constant to the solution

= 6 (e^{x} x - 2 (e^{x} x - e^{x})) + C

\int \frac{1}{16 e ^{- 2 x} + e ^{2 x}} d x

t \to 0 lim (\frac{sin ( k ) t}{t})

\int arccot (15 x) d x

\int 3 x + x^{2} y d x

\frac{1}{10} (e^{5 x} + e^{- 5 x})

What is the integral of 3sqrt(x)e^{sqrt(x)} ?
The integral of 3sqrt(x)e^{sqrt(x)} is 6(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C