What is the integral of (4x)/(sqrt(3x^2-1)) ?

The integral of (4x)/(sqrt(3x^2-1)) is 4/3 sqrt(3x^2-1)+C

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integral of (4x)/(sqrt(3x^2-1))

\int \frac{4 x}{3 x ^{2} - 1} d x

\frac{4}{3} 3 x^{2} - 1 + C

Solution steps

\int \frac{4 x}{3 x ^{2} - 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{3 x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

= 4 \cdot \int \frac{1}{6 u} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{6} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 3 x^{2} - 1

= 4 \cdot \frac{1}{6} \cdot 2 (3 x^{2} - 1)^{\frac{1}{2}}

Simplify

4 \cdot \frac{1}{6} \cdot 2 (3 x^{2} - 1)^{\frac{1}{2}} : \frac{4}{3} 3 x^{2} - 1

= \frac{4}{3} 3 x^{2} - 1

Add a constant to the solution

= \frac{4}{3} 3 x^{2} - 1 + C

\frac{\partial}{\partial x} (e^{- 4 x})

x \to \frac{π}{2} lim (4 sec (x))

\frac{d}{d x} (\frac{4 x}{( x - 1 ) ^{2}})

\int_{0}^{2} 2 x^{2} + 1 d x

What is the integral of (4x)/(sqrt(3x^2-1)) ?
The integral of (4x)/(sqrt(3x^2-1)) is 4/3 sqrt(3x^2-1)+C