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

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

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integral of sqrt(4x^2+4)

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

x 1 + x^{2} + ln x + 1 + x^{2} + C

Solution steps

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

Apply Trigonometric Substitution

: \int 2 sec^{3} (u) d u

= \int 2 sec^{3} (u) d u

Take the constant out:

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

= 2 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

Substitute back

u = arctan (\frac{2}{2} x)

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

Simplify

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

= x 1 + x^{2} + ln x + 1 + x^{2}

Add a constant to the solution

= x 1 + x^{2} + ln x + 1 + x^{2} + C

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