What is the integral of 1/(x^2sqrt(x^2+16)) ?

The integral of 1/(x^2sqrt(x^2+16)) is -(sqrt(16+x^2))/(16x)+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(x^2+16))

\int \frac{1}{x ^{2} x ^{2} + 16} d x

- \frac{16 + x ^{2}}{16 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{16 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{16 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}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{16} (- \frac{1}{v})

Substitute back

= \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})

Simplify

\frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}) : - \frac{16 + x ^{2}}{16 x}

= - \frac{16 + x ^{2}}{16 x}

Add a constant to the solution

= - \frac{16 + x ^{2}}{16 x} + C

y^{^{''}} - 2 y + 1 = 0

x^{2} - 11 y^{2}, y a s

(3 - 5 x)^{9}

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

What is the integral of 1/(x^2sqrt(x^2+16)) ?
The integral of 1/(x^2sqrt(x^2+16)) is -(sqrt(16+x^2))/(16x)+C