What is the integral of 1/(2^x(1+4^x)) ?

The integral of 1/(2^x(1+4^x)) is 1/(ln(2))(-1/(2^x)-arctan(2^x))+C

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integral of 1/(2^x(1+4^x))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{u ^{2} ( 1 + u ^{2} )} : \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1}

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1} d u

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}{ln ( 2 )} (\int \frac{1}{u ^{2}} d u - \int \frac{1}{u ^{2} + 1} d u)

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

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

= \frac{1}{ln ( 2 )} (- \frac{1}{u} - arctan (u))

Substitute back

u = 2^{x}

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

Add a constant to the solution

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

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What is the integral of 1/(2^x(1+4^x)) ?
The integral of 1/(2^x(1+4^x)) is 1/(ln(2))(-1/(2^x)-arctan(2^x))+C