What is the integral of 3/(x^2sqrt(x^2+25)) ?

The integral of 3/(x^2sqrt(x^2+25)) is -(3sqrt(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 3/(x^2sqrt(x^2+25))

\int \frac{3}{x ^{2} x ^{2} + 25} d x

- \frac{3 25 + x ^{2}}{25 x} + C

Solution steps

\int \frac{3}{x ^{2} x ^{2} + 25} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x ^{2} x ^{2} + 25} d x

Apply Trigonometric Substitution

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

= 3 \cdot \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

= 3 \cdot \frac{1}{25} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 3 \cdot \frac{1}{25} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 3 \cdot \frac{1}{25} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 3 \cdot \frac{1}{25} (- \frac{1}{v})

Substitute back

= 3 \cdot \frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )})

Simplify

3 \cdot \frac{1}{25} (- \frac{1}{sin ( arctan ( \frac{1}{5} x ) )}) : - \frac{3 25 + x ^{2}}{25 x}

= - \frac{3 25 + x ^{2}}{25 x}

Add a constant to the solution

= - \frac{3 25 + x ^{2}}{25 x} + C

x \to 0 lim (2 cos (x))

f (x) = 4 x

sin (4 t)

f (x) = cos (2 x) + sin (2 x)

\frac{2.25 r}{( r + 0.5 ) ^{2}}

What is the integral of 3/(x^2sqrt(x^2+25)) ?
The integral of 3/(x^2sqrt(x^2+25)) is -(3sqrt(25+x^2))/(25x)+C