What is the integral of 1/(x^2sqrt(25+x^2)) ?

The integral of 1/(x^2sqrt(25+x^2)) is -(sqrt(25+x^2))/(25x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(25+x^2))

\int \frac{1}{x ^{2} 25 + x ^{2}} d x

- \frac{25 + x ^{2}}{25 x} + C

Solution steps

\int \frac{1}{x ^{2} 25 + x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{25 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{25 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}{25} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{25} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{25} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{25} (- \frac{1}{v})

Substitute back

= \frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )})

Simplify

\frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )}) : - \frac{25 + x ^{2}}{25 x}

= - \frac{25 + x ^{2}}{25 x}

Add a constant to the solution

= - \frac{25 + x ^{2}}{25 x} + C

\frac{d}{d x} (sin^{2} (x) - sin (x^{2}))

5 t^{3}

(4, 1), (6, 7)

\int sin^{2} (x) sin (x) d x

\int_{1}^{10} \frac{1}{( x - 2 ) ^{\frac{1}{3}}} d x

What is the integral of 1/(x^2sqrt(25+x^2)) ?
The integral of 1/(x^2sqrt(25+x^2)) is -(sqrt(25+x^2))/(25x)+C