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

The integral of sqrt(4t^2+4) is tsqrt(1+t^2)+ln|t+sqrt(1+t^2)|+C

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integral of sqrt(4t^2+4)

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

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

Solution steps

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

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} t)

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

Simplify

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

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

Add a constant to the solution

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

\int sin (2 x + π) d x

\frac{d y}{d x} = \frac{e ^{x} - y}{x}

x \to 0 lim (x sin (\frac{1}{x}))

- 5 t^{2} + 75 t - 30

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