What is the integral of 2^{x/2} ?

The integral of 2^{x/2} is (2^{1/2 (x+2)})/(ln(2))+C

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integral of 2^{x/2}

\int 2^{\frac{x}{2}} d x

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

Solution steps

\int 2^{\frac{x}{2}} d x

Apply u-substitution

: \int 2^{u + 1} d u

= \int 2^{u + 1} d u

2^{u + 1} = 2^{1} \cdot 2^{u}

= \int 2^{1} \cdot 2^{u} d u

Take the constant out:

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

= 2^{1} \cdot \int 2^{u} d u

Simplify

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

\int a^{x} d x = \frac{a ^{x}}{ln a}

= 2 \cdot \frac{2 ^{u}}{ln ( 2 )}

Substitute back

u = \frac{x}{2}

= 2 \cdot \frac{2 ^{\frac{x}{2}}}{ln ( 2 )}

Simplify

2 \cdot \frac{2 ^{\frac{x}{2}}}{ln ( 2 )} : \frac{2 ^{\frac{1}{2} (x + 2)}}{ln ( 2 )}

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

Add a constant to the solution

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

x \cdot \frac{d y}{d x} = 2 x e^{x} - y + 6 x^{2}

\frac{d}{d y} (- x y sin (x) + 2 y cos (x))

y^{^{''}} + y = tan^{2} (x)

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = x^{2}

f (x) = \frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

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