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

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

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integral of sqrt(2+4t^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

= \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) ∣

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

Substitute back

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

= \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

\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)) : \frac{1}{2} t 1 + 2 t^{2} + \frac{1}{2} ln 2 t + 1 + 2 t^{2}

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

Add a constant to the solution

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

\int x^{3} 3^{x} d x

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9 x^{2}

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