What is the integral of (2t)/(sqrt(1-4t^4)) ?

The integral of (2t)/(sqrt(1-4t^4)) is 1/2 arcsin(2t^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2t)/(sqrt(1-4t^4))

\int \frac{2 t}{1 - 4 t ^{4}} d t

\frac{1}{2} arcsin (2 t^{2}) + C

Solution steps

\int \frac{2 t}{1 - 4 t ^{4}} d t

Take the constant out:

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

= 2 \cdot \int \frac{t}{1 - 4 t ^{4}} d t

Apply u-substitution

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

= 2 \cdot \int \frac{1}{2 1 - 4 u ^{2}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 4 u ^{2}} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

= \frac{1}{2} arcsin (2 t^{2})

Add a constant to the solution

= \frac{1}{2} arcsin (2 t^{2}) + C

\frac{d x}{d t} = - x

\frac{d}{d x} (1.5)

\frac{d}{d x} (- 2 x^{2} + 3 x - 6)

f (x) = 7 e^{x} cos (x)

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

What is the integral of (2t)/(sqrt(1-4t^4)) ?
The integral of (2t)/(sqrt(1-4t^4)) is 1/2 arcsin(2t^2)+C