What is the integral of (600x)/(sqrt(x^2+49)) ?

The integral of (600x)/(sqrt(x^2+49)) is 600sqrt(x^2+49)+C

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integral of (600x)/(sqrt(x^2+49))

\int \frac{600 x}{x ^{2} + 49} d x

600 x^{2} + 49 + C

Solution steps

\int \frac{600 x}{x ^{2} + 49} d x

Take the constant out:

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

= 600 \cdot \int \frac{x}{x ^{2} + 49} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 600 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 600 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 600 \cdot \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 49

= 600 \cdot \frac{1}{2} \cdot 2 (x^{2} + 49)^{\frac{1}{2}}

Simplify

600 \cdot \frac{1}{2} \cdot 2 (x^{2} + 49)^{\frac{1}{2}} : 600 x^{2} + 49

= 600 x^{2} + 49

Add a constant to the solution

= 600 x^{2} + 49 + C

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What is the integral of (600x)/(sqrt(x^2+49)) ?
The integral of (600x)/(sqrt(x^2+49)) is 600sqrt(x^2+49)+C