What is the integral of sqrt(1+8t^2) ?

The integral of sqrt(1+8t^2) is 1/2 tsqrt(8t^2+1)+1/(4sqrt(2))ln|2sqrt(2)t+sqrt(1+8t^2)|+C

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integral of sqrt(1+8t^2)

\int 1 + 8 t^{2} d t

\frac{1}{2} t 8 t^{2} + 1 + \frac{1}{4 2} ln 22 t + 1 + 8 t^{2} + C

Solution steps

\int 1 + 8 t^{2} d t

Apply Trigonometric Substitution

: \int \frac{1}{2 2} sec^{3} (u) d u

= \int \frac{1}{2 2} sec^{3} (u) d u

Take the constant out:

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

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

Apply Integral Reduction

= \frac{1}{2 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) ∣

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

Substitute back

u = arctan (22 t)

= \frac{1}{2 2} (\frac{sec ^{2} ( arctan ( 2 2 t ) ) sin ( arctan ( 2 2 t ) )}{2} + \frac{1}{2} ln tan (arctan (22 t)) + sec (arctan (22 t)))

Simplify

\frac{1}{2 2} (\frac{sec ^{2} ( arctan ( 2 2 t ) ) sin ( arctan ( 2 2 t ) )}{2} + \frac{1}{2} ln tan (arctan (22 t)) + sec (arctan (22 t))) : \frac{1}{2} t 8 t^{2} + 1 + \frac{1}{4 2} ln 22 t + 1 + 8 t^{2}

= \frac{1}{2} t 8 t^{2} + 1 + \frac{1}{4 2} ln 22 t + 1 + 8 t^{2}

Add a constant to the solution

= \frac{1}{2} t 8 t^{2} + 1 + \frac{1}{4 2} ln 22 t + 1 + 8 t^{2} + C

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What is the integral of sqrt(1+8t^2) ?
The integral of sqrt(1+8t^2) is 1/2 tsqrt(8t^2+1)+1/(4sqrt(2))ln|2sqrt(2)t+sqrt(1+8t^2)|+C