What is the integral of wsqrt(w^2-ln(3)) ?

The integral of wsqrt(w^2-ln(3)) is 1/3 (w^2-ln(3))^{3/2}+C

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integral of wsqrt(w^2-ln(3))

\int w w^{2} - ln (3) d w

\frac{1}{3} (w^{2} - ln (3))^{\frac{3}{2}} + C

Solution steps

\int w w^{2} - ln (3) d w

Apply u-substitution

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

= \int \frac{u}{2} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = w^{2} - ln (3)

= \frac{1}{2} \cdot \frac{2}{3} (w^{2} - ln (3))^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{2}{3} (w^{2} - ln (3))^{\frac{3}{2}} : \frac{1}{3} (w^{2} - ln (3))^{\frac{3}{2}}

= \frac{1}{3} (w^{2} - ln (3))^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{3} (w^{2} - ln (3))^{\frac{3}{2}} + C

x \to 0 lim (\frac{3 x}{x e ^{x}})

\int \frac{3 x ^{2} - 2 x + 5}{x ^{2}} d x

x^{3} \frac{d y}{d x} + y^{2} = x^{2} y

\frac{3 x ^{2} + 1}{2 x ^{2} - 5}

f (x) = x^{3} x

What is the integral of wsqrt(w^2-ln(3)) ?
The integral of wsqrt(w^2-ln(3)) is 1/3 (w^2-ln(3))^{3/2}+C