What is the integral of sqrt(2x^2+9) ?

The integral of sqrt(2x^2+9) is 1/2 xsqrt(9+2x^2)+1/2*9/(sqrt(2))ln((|sqrt(2)x+sqrt(9+2x^2)|)/3)+C

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integral of sqrt(2x^2+9)

\int 2 x^{2} + 9 d x

\frac{1}{2} x 9 + 2 x^{2} + \frac{1}{2} \cdot \frac{9}{2} ln (\frac{2 x + 9 + 2 x ^{2}}{3}) + C

Solution steps

\int 2 x^{2} + 9 d x

Apply Trigonometric Substitution

: \int \frac{9}{2} sec^{3} (u) d u

= \int \frac{9}{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{9}{2} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{9}{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{9}{2} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{2}{3} x)

= \frac{9}{2} \frac{sec ^{2} ( arctan ( \frac{2}{3} x ) ) sin ( arctan ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x))

Simplify

\frac{9}{2} \frac{sec ^{2} ( arctan ( \frac{2}{3} x ) ) sin ( arctan ( \frac{2}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} x)) + sec (arctan (\frac{2}{3} x)) : \frac{1}{2} x 9 + 2 x^{2} + \frac{1}{2} \cdot \frac{9}{2} ln (\frac{2 x + 9 + 2 x ^{2}}{3})

= \frac{1}{2} x 9 + 2 x^{2} + \frac{1}{2} \cdot \frac{9}{2} ln (\frac{2 x + 9 + 2 x ^{2}}{3})

Add a constant to the solution

= \frac{1}{2} x 9 + 2 x^{2} + \frac{1}{2} \cdot \frac{9}{2} ln (\frac{2 x + 9 + 2 x ^{2}}{3}) + C

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What is the integral of sqrt(2x^2+9) ?
The integral of sqrt(2x^2+9) is 1/2 xsqrt(9+2x^2)+1/2*9/(sqrt(2))ln((|sqrt(2)x+sqrt(9+2x^2)|)/3)+C