What is the integral of b^{4x} ?

The integral of b^{4x} is (b^{4x})/(4ln(b))+C

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integral of b^{4x}

\int b^{4 x} d x

\frac{b ^{4 x}}{4 ln ( b )} + C

Solution steps

\int b^{4 x} d x

Apply u-substitution

: \int b^{u} \frac{1}{4} d u

= \int b^{u} \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int b^{u} d u

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

= \frac{1}{4} \cdot \frac{b ^{u}}{ln ( b )}

Substitute back

u = 4 x

= \frac{1}{4} \cdot \frac{b ^{4 x}}{ln ( b )}

Simplify

\frac{1}{4} \cdot \frac{b ^{4 x}}{ln ( b )} : \frac{b ^{4 x}}{4 ln ( b )}

= \frac{b ^{4 x}}{4 ln ( b )}

Add a constant to the solution

= \frac{b ^{4 x}}{4 ln ( b )} + C

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

\frac{\partial}{\partial x} ((x^{2} - y^{2})^{0.5})

\int x 2 x + 1 d x

\frac{\partial}{\partial y} (- \frac{sin ( 2 x y )}{x y})

x^{- \frac{3}{4}} + x^{\frac{3}{4}}