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

The integral of (5tsqrt(t))/(1+t^5) is 2arctan(t^{5/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5tsqrt(t))/(1+t^5)

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

2 arctan (t^{\frac{5}{2}}) + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{t t}{1 + t ^{5}} d t

Apply u-substitution

: \int \frac{2 u ^{4}}{1 + u ^{10}} d u

= 5 \cdot \int \frac{2 u ^{4}}{1 + u ^{10}} d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int \frac{u ^{4}}{1 + u ^{10}} d u

Apply u-substitution

: \int \frac{1}{5 ( 1 + v ^{2} )} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= 5 \cdot 2 \cdot \frac{1}{5} arctan (v)

Substitute back

= 5 \cdot 2 \cdot \frac{1}{5} arctan ((t)^{5})

Simplify

5 \cdot 2 \cdot \frac{1}{5} arctan ((t)^{5}) : 2 arctan (t^{\frac{5}{2}})

= 2 arctan (t^{\frac{5}{2}})

Add a constant to the solution

= 2 arctan (t^{\frac{5}{2}}) + C

y = sin (x) \cdot cos (x)

f (x) = x - 2 ln (1 - \frac{1}{x})

27 t^{2} sin (6 t - 6 0^{\circ})

\frac{d}{d x} (x^{2} + y^{2} - 5)

\frac{d}{d x} (3 x - \frac{1}{x})

What is the integral of (5tsqrt(t))/(1+t^5) ?
The integral of (5tsqrt(t))/(1+t^5) is 2arctan(t^{5/2})+C