What is the integral of (x^2)/(sqrt(x^6+36)) ?

The integral of (x^2)/(sqrt(x^6+36)) is 1/3 ln((|x^3+sqrt(36+x^6)|)/6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(x^6+36))

\int \frac{x ^{2}}{x ^{6} + 36} d x

\frac{1}{3} ln (\frac{x ^{3} + 36 + x ^{6}}{6}) + C

Solution steps

\int \frac{x ^{2}}{x ^{6} + 36} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 36} d u

= \int \frac{1}{3 u ^{2} + 36} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{2} + 36} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{3} ln tan (arctan (\frac{1}{6} x^{3})) + sec (arctan (\frac{1}{6} x^{3}))

Simplify

\frac{1}{3} ln tan (arctan (\frac{1}{6} x^{3})) + sec (arctan (\frac{1}{6} x^{3})) : \frac{1}{3} ln (\frac{x ^{3} + 36 + x ^{6}}{6})

= \frac{1}{3} ln (\frac{x ^{3} + 36 + x ^{6}}{6})

Add a constant to the solution

= \frac{1}{3} ln (\frac{x ^{3} + 36 + x ^{6}}{6}) + C

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

\int 8 cos^{3} (5 x) d x

\int cot (2 x) (cot (2 x) + 3 csc (2 x)) d x

\int (1 - x)^{\frac{- 1}{2}} d x

\int \frac{e ^{- 2 t}}{( 1 + e ^{- t} )} d t

What is the integral of (x^2)/(sqrt(x^6+36)) ?
The integral of (x^2)/(sqrt(x^6+36)) is 1/3 ln((|x^3+sqrt(36+x^6)|)/6)+C