What is the integral of (2^x-2)^2 ?

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

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integral of (2^x-2)^2

\int (2^{x} - 2)^{2} d x

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

Solution steps

\int (2^{x} - 2)^{2} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{( v - 2 ) ^{2}}{v} d v

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

Expand

\frac{( v - 2 ) ^{2}}{v} : v - 4 + \frac{4}{v}

= \frac{1}{ln ( 2 )} \cdot \int v - 4 + \frac{4}{v} d v

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 v d v - \int 4 d v + \int \frac{4}{v} d v)

\int v d v = \frac{v ^{2}}{2}

\int 4 d v = 4 v

\int \frac{4}{v} d v = 4 ln ∣ v ∣

= \frac{1}{ln ( 2 )} (\frac{v ^{2}}{2} - 4 v + 4 ln ∣ v ∣)

Substitute back

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

Simplify

\frac{1}{ln ( 2 )} (\frac{( 2 ^{x} - 2 + 2 ) ^{2}}{2} - 4 (2^{x} - 2 + 2) + 4 ln ∣ 2^{x} - 2 + 2 ∣) : \frac{2 ^{2 x - 1} - 2 ^{2 + x} + 4 ln ( 2 ) x}{ln ( 2 )}

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

Add a constant to the solution

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

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