What is the integral of (3x)/(sqrt(1-16x^4)) ?

The integral of (3x)/(sqrt(1-16x^4)) is 3/8 arcsin(4x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x)/(sqrt(1-16x^4))

\int \frac{3 x}{1 - 16 x ^{4}} d x

\frac{3}{8} arcsin (4 x^{2}) + C

Solution steps

\int \frac{3 x}{1 - 16 x ^{4}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{1 - 16 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 1 - 16 u ^{2}} d u

= 3 \cdot \int \frac{1}{2 1 - 16 u ^{2}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{1 - 16 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 x^{2})

Simplify

3 \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (4 x^{2}) : \frac{3}{8} arcsin (4 x^{2})

= \frac{3}{8} arcsin (4 x^{2})

Add a constant to the solution

= \frac{3}{8} arcsin (4 x^{2}) + C

\int x (7 x^{2} + 5)^{9} d x

\int \frac{e ^{2 x}}{e ^{2 x} + 8} d x

\int - 3 sin (x) + 5 cos (x) d x

\int (6 x^{2} + \frac{2}{x ^{2}} + 5) d x

\int e^{t} \frac{1}{t} d t

What is the integral of (3x)/(sqrt(1-16x^4)) ?
The integral of (3x)/(sqrt(1-16x^4)) is 3/8 arcsin(4x^2)+C