What is the integral of 1/(x^2sqrt(x^2+100)) ?

The integral of 1/(x^2sqrt(x^2+100)) is -(sqrt(100+x^2))/(100x)+C

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integral of 1/(x^2sqrt(x^2+100))

\int \frac{1}{x ^{2} x ^{2} + 100} d x

- \frac{100 + x ^{2}}{100 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} + 100} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{100 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{100 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}{100} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{100} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{100} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{100} (- \frac{1}{v})

Substitute back

= \frac{1}{100} (- \frac{1}{sin ( arctan ( \frac{1}{10} x ) )})

Simplify

\frac{1}{100} (- \frac{1}{sin ( arctan ( \frac{1}{10} x ) )}) : - \frac{100 + x ^{2}}{100 x}

= - \frac{100 + x ^{2}}{100 x}

Add a constant to the solution

= - \frac{100 + x ^{2}}{100 x} + C

f (x) = x^{8} - 9 e^{3 x}

\int_{0}^{π} cos (t) d t

\frac{d}{d x} (2 x^{3} - 12 x^{2} + 18 x)

x \to 3 lim ((- 4) x)

\frac{d x}{d y} + 800 y = 400

What is the integral of 1/(x^2sqrt(x^2+100)) ?
The integral of 1/(x^2sqrt(x^2+100)) is -(sqrt(100+x^2))/(100x)+C