What is the integral of (e^{3x}+1)^2 ?

The integral of (e^{3x}+1)^2 is 1/6 e^{6x}+2/3 e^{3x}+x+C

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integral of (e^{3x}+1)^2

\int (e^{3 x} + 1)^{2} d x

\frac{1}{6} e^{6 x} + \frac{2}{3} e^{3 x} + x + C

Solution steps

\int (e^{3 x} + 1)^{2} d x

Apply u-substitution

: \int \frac{( e ^{u} + 1 ) ^{2}}{3} d u

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

Take the constant out:

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

= \frac{1}{3} \cdot \int (e^{u} + 1)^{2} d u

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{( v + 1 ) ^{2}}{v} d v

Expand

\frac{( v + 1 ) ^{2}}{v} : v + 2 + \frac{1}{v}

= \frac{1}{3} \cdot \int v + 2 + \frac{1}{v} d v

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}{3} (\int v d v + \int 2 d v + \int \frac{1}{v} d v)

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

= \frac{1}{3} (\frac{v ^{2}}{2} + 2 v + ln ∣ v ∣)

Substitute back

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

Simplify

\frac{1}{3} (\frac{( e ^{3 x} ) ^{2}}{2} + 2 e^{3 x} + ln e^{3 x}) : \frac{1}{6} e^{6 x} + \frac{2}{3} e^{3 x} + x

= \frac{1}{6} e^{6 x} + \frac{2}{3} e^{3 x} + x

Add a constant to the solution

= \frac{1}{6} e^{6 x} + \frac{2}{3} e^{3 x} + x + C

y^{^{''}} + 2 y^{^{'}} - 3 = 0

x \to - 5 - lim (\frac{3 x}{2 x + 10})

x \to 0 - lim ((\frac{1}{5})^{\frac{1}{x}})

x \to 1 - 0 lim (\frac{1}{1 - x})

x \to \infty + lim (x ln (x))

What is the integral of (e^{3x}+1)^2 ?
The integral of (e^{3x}+1)^2 is 1/6 e^{6x}+2/3 e^{3x}+x+C