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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x^{2} y^{^{'}} + 2 x y = 7 cos^{2} (x)

\int (\frac{sin ( 2 x )}{cos ( x )}) d x

\frac{d}{d x} (\frac{x ^{6}}{5})

\frac{d}{d x} (\frac{y}{x + y})

f (x) = - \frac{1}{7} (x^{- 7} - x^{14})

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