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

The integral of e^{4x}sqrt(1+e^{2x)} is 1/15 (3e^{2x}-2)(e^{2x}+1)^{3/2}+C

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

\int e^{4 x} 1 + e^{2 x} d x

\frac{1}{15} (3 e^{2 x} - 2) (e^{2 x} + 1)^{\frac{3}{2}} + C

Solution steps

\int e^{4 x} 1 + e^{2 x} d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{2} d u

= \int \frac{u ( u - 1 )}{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 u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

= \frac{1}{2} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + e^{2 x}

= \frac{1}{2} (\frac{2}{5} (1 + e^{2 x})^{\frac{5}{2}} - \frac{2}{3} (1 + e^{2 x})^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{5} (1 + e^{2 x})^{\frac{5}{2}} - \frac{2}{3} (1 + e^{2 x})^{\frac{3}{2}}) : \frac{1}{15} (3 e^{2 x} - 2) (e^{2 x} + 1)^{\frac{3}{2}}

= \frac{1}{15} (3 e^{2 x} - 2) (e^{2 x} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 e^{2 x} - 2) (e^{2 x} + 1)^{\frac{3}{2}} + C

x \to 3 lim (2 x^{2} + 4 x + 1)

y^{^{''}} - 2 y^{^{'}} + 5 y = 25 x^{2} + 12

\int_{0}^{2} \frac{3 t}{( 3 - t ) ^{2}} d t

y = 3 x^{2} + 5 x, (- 3, - 13)

x \to 3 lim (e^{x})

What is the integral of e^{4x}sqrt(1+e^{2x)} ?
The integral of e^{4x}sqrt(1+e^{2x)} is 1/15 (3e^{2x}-2)(e^{2x}+1)^{3/2}+C