What is the integral of (19x)/(sqrt(2x^2+6)) ?

The integral of (19x)/(sqrt(2x^2+6)) is 19/2 sqrt(2x^2+6)+C

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integral of (19x)/(sqrt(2x^2+6))

\int \frac{19 x}{2 x ^{2} + 6} d x

\frac{19}{2} 2 x^{2} + 6 + C

Solution steps

\int \frac{19 x}{2 x ^{2} + 6} d x

Take the constant out:

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

= 19 \cdot \int \frac{x}{2 x ^{2} + 6} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 19 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 19 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 19 \cdot \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 2 x^{2} + 6

= 19 \cdot \frac{1}{4} \cdot 2 (2 x^{2} + 6)^{\frac{1}{2}}

Simplify

19 \cdot \frac{1}{4} \cdot 2 (2 x^{2} + 6)^{\frac{1}{2}} : \frac{19}{2} 2 x^{2} + 6

= \frac{19}{2} 2 x^{2} + 6

Add a constant to the solution

= \frac{19}{2} 2 x^{2} + 6 + C

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What is the integral of (19x)/(sqrt(2x^2+6)) ?
The integral of (19x)/(sqrt(2x^2+6)) is 19/2 sqrt(2x^2+6)+C