What is the solution for f^'(x)=a^{2x} ?

The solution for f^'(x)=a^{2x} is f(x)=a^{2x}x-(2a^{2x}xln(a)-a^{2x})/(2ln(a))+c_{1}

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f^'(x)=a^{2x}

f^{^{'}} (x) = a^{2 x}

f (x) = a^{2 x} x - \frac{2 a ^{2 x} x ln ( a ) - a ^{2 x}}{2 ln ( a )} + c_{1}

Solution steps

f^{'} (x) = a^{2 x}

f^{^{'}} (x) = g (x)

then

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

f (x) = \int a^{2 x} d x

\int a^{2 x} d x = a^{2 x} x - \frac{2 a ^{2 x} x ln ( a ) - a ^{2 x}}{2 ln ( a )} + c_{1}

f (x) = a^{2 x} x - \frac{2 a ^{2 x} x ln ( a ) - a ^{2 x}}{2 ln ( a )} + c_{1}

n = 1 \sum \infty (\frac{n - 2}{n})^{n}

\int (sin (2 u) - cos (3 u))^{(2)} d u

\frac{d}{d x} (- 3 x - 9)

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

[x + (x + sin^{2} (x))^{3}]^{4}

What is the solution for f^'(x)=a^{2x} ?
The solution for f^'(x)=a^{2x} is f(x)=a^{2x}x-(2a^{2x}xln(a)-a^{2x})/(2ln(a))+c_{1}