What is the integral of x/(1000)sqrt(x^2+2500) ?

The integral of x/(1000)sqrt(x^2+2500) is 1/3000 (x^2+2500)^{3/2}+C

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integral of x/(1000)sqrt(x^2+2500)

\int \frac{x}{1000} x^{2} + 2500 d x

\frac{1}{3000} (x^{2} + 2500)^{\frac{3}{2}} + C

Solution steps

\int \frac{x}{1000} x^{2} + 2500 d x

Take the constant out:

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

= \frac{1}{1000} \cdot \int x x^{2} + 2500 d x

Apply u-substitution

: \int u^{2} d u

= \frac{1}{1000} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{1}{1000} \cdot \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 2500

= \frac{1}{1000} \cdot \frac{( x ^{2} + 2500 ) ^{3}}{3}

Simplify

\frac{1}{1000} \cdot \frac{( x ^{2} + 2500 ) ^{3}}{3} : \frac{1}{3000} (x^{2} + 2500)^{\frac{3}{2}}

= \frac{1}{3000} (x^{2} + 2500)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3000} (x^{2} + 2500)^{\frac{3}{2}} + C

\frac{d}{d x} ((x - 5) \cdot ln (\frac{x}{5}))

\frac{d}{d x} (- \frac{375}{x ^{4}})

\int_{- 2}^{3} x^{2} - x - 6 d x

300 x - \frac{1}{2} x^{2}

x \to 6 lim (x + 2)

What is the integral of x/(1000)sqrt(x^2+2500) ?
The integral of x/(1000)sqrt(x^2+2500) is 1/3000 (x^2+2500)^{3/2}+C