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

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

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

\int 2^{x} 1 + 4^{x} d x

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

Solution steps

\int 2^{x} 1 + 4^{x} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \frac{1}{ln ( 2 )} \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{ln ( 2 )} (\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) ∣

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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