What is the integral of 1/(t^2sqrt(t^2+5)) ?

The integral of 1/(t^2sqrt(t^2+5)) is -(sqrt(5+t^2))/(5t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(t^2sqrt(t^2+5))

\int \frac{1}{t ^{2} t ^{2} + 5} d t

- \frac{5 + t ^{2}}{5 t} + C

Solution steps

\int \frac{1}{t ^{2} t ^{2} + 5} d t

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5 tan ^{2} ( u )} d u

= \int \frac{sec ( u )}{5 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}{5} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{5} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{1}{5} (- \frac{1}{v})

Substitute back

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

Simplify

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

= - \frac{5 + t ^{2}}{5 t}

Add a constant to the solution

= - \frac{5 + t ^{2}}{5 t} + C

\frac{d}{d x} (sin (x^{2} + π))

y = x^{x} \frac{d y}{d x}

\frac{d}{d x} (\frac{e ^{x} + e ^{- x}}{2})

\int e^{y} \cdot y^{2} d y

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

What is the integral of 1/(t^2sqrt(t^2+5)) ?
The integral of 1/(t^2sqrt(t^2+5)) is -(sqrt(5+t^2))/(5t)+C