What is the integral of sqrt(5+2x^2) ?

The integral of sqrt(5+2x^2) is 1/2 xsqrt(5+2x^2)+5/(2sqrt(2))ln|sqrt(2/5)x+sqrt(1/5 (5+2x^2))|+C

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integral of sqrt(5+2x^2)

\int 5 + 2 x^{2} d x

\frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2}) + C

Solution steps

\int 5 + 2 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{5}{2} sec^{3} (u) d u

= \int \frac{5}{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{5}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{5}{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{5}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{5} x)

= \frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x))

Simplify

\frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x)) : \frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2})

= \frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2})

Add a constant to the solution

= \frac{1}{2} x 5 + 2 x^{2} + \frac{5}{2 2} ln \frac{2}{5} x + \frac{1}{5} (5 + 2 x^{2}) + C

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What is the integral of sqrt(5+2x^2) ?
The integral of sqrt(5+2x^2) is 1/2 xsqrt(5+2x^2)+5/(2sqrt(2))ln|sqrt(2/5)x+sqrt(1/5 (5+2x^2))|+C