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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{3 sec ( u )}{4 tan ^{2} ( u )} d u

= \int \frac{3 sec ( u )}{4 tan ^{2} ( u )} d u

Take the constant out:

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

= \frac{3}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{3}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{3}{4} (- \frac{1}{v})

Substitute back

= \frac{3}{4} - \frac{1}{sin ( arctan ( \frac{3}{2} x ) )}

Simplify

\frac{3}{4} - \frac{1}{sin ( arctan ( \frac{3}{2} x ) )} : - \frac{4 + 3 x ^{2}}{4 x}

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

Add a constant to the solution

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

y = 1 n (x^{3})

\frac{\partial}{\partial y} (\frac{x}{x ^{2} + y ^{2} + z ^{2}})

y = 2 + ∣ x - 1 ∣, y = - \frac{1}{5} x + 7

\int_{1}^{5 e} \frac{1}{3 x} d x

\int \frac{5 ln ( x )}{x} d x

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