What is the integral of (2x)/(sqrt(x^4+36)) ?

The integral of (2x)/(sqrt(x^4+36)) is ln(1/6 |x^2+sqrt(36+x^4)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x)/(sqrt(x^4+36))

\int \frac{2 x}{x ^{4} + 36} d x

ln (\frac{1}{6} x^{2} + 36 + x^{4}) + C

Solution steps

\int \frac{2 x}{x ^{4} + 36} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x ^{4} + 36} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 36} d u

= 2 \cdot \int \frac{1}{2 u ^{2} + 36} 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}{u ^{2} + 36} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{1}{6} x^{2})) + sec (arctan (\frac{1}{6} x^{2}))

Simplify

2 \cdot \frac{1}{2} ln tan (arctan (\frac{1}{6} x^{2})) + sec (arctan (\frac{1}{6} x^{2})) : ln (\frac{1}{6} x^{2} + 36 + x^{4})

= ln (\frac{1}{6} x^{2} + 36 + x^{4})

Add a constant to the solution

= ln (\frac{1}{6} x^{2} + 36 + x^{4}) + C

\frac{1}{s} \cdot (e^{- 2 \cdot s \cdot π} - 1)

\frac{d P}{d t} = (P - P^{2})

\int_{4}^{2} (6 x^{2} - 2 x + 1) d x

\frac{\partial}{\partial x} (- 3 y (- 2 x - y - 9))

(2 t - 1)^{2}

What is the integral of (2x)/(sqrt(x^4+36)) ?
The integral of (2x)/(sqrt(x^4+36)) is ln(1/6 |x^2+sqrt(36+x^4)|)+C