What is the integral of 1/(x^2sqrt(x^2+36)) ?

The integral of 1/(x^2sqrt(x^2+36)) is -(sqrt(36+x^2))/(36x)+C

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integral of 1/(x^2sqrt(x^2+36))

\int \frac{1}{x ^{2} x ^{2} + 36} d x

- \frac{36 + x ^{2}}{36 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} + 36} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{36 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{36 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}{36} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{36} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{36} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{36} (- \frac{1}{v})

Substitute back

= \frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )})

Simplify

\frac{1}{36} (- \frac{1}{sin ( arctan ( \frac{1}{6} x ) )}) : - \frac{36 + x ^{2}}{36 x}

= - \frac{36 + x ^{2}}{36 x}

Add a constant to the solution

= - \frac{36 + x ^{2}}{36 x} + C

x \to 2 lim (x^{3} + x^{2} - 5 x + 1)

\int \frac{x + 1}{x ^{5}} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 20 y = 0

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

\frac{d}{d x} (3^{5 x})

What is the integral of 1/(x^2sqrt(x^2+36)) ?
The integral of 1/(x^2sqrt(x^2+36)) is -(sqrt(36+x^2))/(36x)+C