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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{a ^{2}} (- \frac{1}{v})

Substitute back

= \frac{1}{a ^{2}} (- \frac{1}{sin ( arctan ( \frac{1}{a} x ) )})

Simplify

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

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

Add a constant to the solution

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

e^{- t}

\frac{d}{d x} (c x^{- 4})

\int (2 x^{2} + 9) (x - 3)^{8} d x

\frac{\partial}{\partial x} (x^{2} - y^{2})

\frac{\partial}{\partial y} (ln (1 + x^{2} + y^{2}))

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