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

The integral of e^xsqrt(e^{2x)+4} is 4(1/8 e^xsqrt(4+e^{2x)}+1/2 ln(1/2 |e^x+sqrt(4+e^{2x)}|))+C

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

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

4 (\frac{1}{8} e^{x} 4 + e^{2 x} + \frac{1}{2} ln (\frac{1}{2} e^{x} + 4 + e^{2 x})) + C

Solution steps

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

Apply u-substitution

: \int u^{2} + 4 d u

= \int u^{2} + 4 d u

Apply Trigonometric Substitution

: \int 4 sec^{3} (v) d v

= \int 4 sec^{3} (v) d v

Take the constant out:

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

= 4 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} e ^{x} ) ) sin ( arctan ( \frac{1}{2} e ^{x} ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} e^{x})) + sec (arctan (\frac{1}{2} e^{x})))

Simplify

4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} e ^{x} ) ) sin ( arctan ( \frac{1}{2} e ^{x} ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} e^{x})) + sec (arctan (\frac{1}{2} e^{x}))) : 4 (\frac{1}{8} e^{x} 4 + e^{2 x} + \frac{1}{2} ln (\frac{1}{2} e^{x} + 4 + e^{2 x}))

= 4 (\frac{1}{8} e^{x} 4 + e^{2 x} + \frac{1}{2} ln (\frac{1}{2} e^{x} + 4 + e^{2 x}))

Add a constant to the solution

= 4 (\frac{1}{8} e^{x} 4 + e^{2 x} + \frac{1}{2} ln (\frac{1}{2} e^{x} + 4 + e^{2 x})) + C

\int cos^{4} (x) \cdot sin^{3} (x) d x

\frac{d}{d x} (x^{2} + sin (2 x))

\frac{d}{d x} (\frac{3 ( 2 x + 1 )}{6 x + 6 x ^{2}})

\frac{d y}{d x} = sin (3 y) ln (x)

sin (x) + \frac{7}{2} cot (x)

What is the integral of e^xsqrt(e^{2x)+4} ?
The integral of e^xsqrt(e^{2x)+4} is 4(1/8 e^xsqrt(4+e^{2x)}+1/2 ln(1/2 |e^x+sqrt(4+e^{2x)}|))+C