What is the integral of (x^2+y^2)^{3/2} ?

The integral of (x^2+y^2)^{3/2} is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+y^2)^{3/2}

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

\frac{1}{4} x (x^{2} + y^{2})^{\frac{3}{2}} + \frac{3 y ^{2} x ( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}{8} + \frac{3 y ^{4}}{8} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣} + C

Solution steps

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

y^{2} + x^{2} = y^{2} (1 + \frac{x ^{2}}{y ^{2}})

= \int (y^{2} (1 + \frac{x ^{2}}{y ^{2}}))^{\frac{3}{2}} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{y ^{2}} = (\frac{x}{y ^{2}})^{2}

= \int y^{2} 1 + (\frac{x}{y ^{2}})^{2}^{\frac{3}{2}} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

y^{2} 1 + (\frac{x}{y ^{2}})^{2}^{\frac{3}{2}} = (y^{2})^{\frac{3}{2}} 1 + (\frac{x}{y ^{2}})^{2}^{\frac{3}{2}}

= \int (y^{2})^{\frac{3}{2}} 1 + (\frac{x}{y ^{2}})^{2}^{\frac{3}{2}} d x

Take the constant out:

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

= (y^{2})^{\frac{3}{2}} \cdot \int 1 + (\frac{x}{y ^{2}})^{2}^{\frac{3}{2}} d x

Apply Trigonometric Substitution

: \int y sec^{5} (u) d u

= (y^{2})^{\frac{3}{2}} \cdot \int y sec^{5} (u) d u

Take the constant out:

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

= (y^{2})^{\frac{3}{2}} y \cdot \int sec^{5} (u) d u

Apply Integral Reduction

= (y^{2})^{\frac{3}{2}} y (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} \cdot \int sec^{3} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= (y^{2})^{\frac{3}{2}} y (\frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

Substitute back

u = arctan (\frac{1}{y} x)

= (y^{2})^{\frac{3}{2}} y \frac{sec ^{4} ( arctan ( \frac{1}{y} x ) ) sin ( arctan ( \frac{1}{y} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{y} x)) tan (arctan (\frac{1}{y} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{y} x)) + sec (arctan (\frac{1}{y} x)))

Simplify

(y^{2})^{\frac{3}{2}} y \frac{sec ^{4} ( arctan ( \frac{1}{y} x ) ) sin ( arctan ( \frac{1}{y} x ) )}{4} + \frac{3}{4} (\frac{1}{2} sec (arctan (\frac{1}{y} x)) tan (arctan (\frac{1}{y} x)) + \frac{1}{2} ln tan (arctan (\frac{1}{y} x)) + sec (arctan (\frac{1}{y} x))) : \frac{1}{4} x (x^{2} + y^{2})^{\frac{3}{2}} + \frac{3 y ^{2} x ( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}{8} + \frac{3 y ^{4}}{8} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣}

= \frac{1}{4} x (x^{2} + y^{2})^{\frac{3}{2}} + \frac{3 y ^{2} x ( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}{8} + \frac{3 y ^{4}}{8} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣}

Add a constant to the solution

= \frac{1}{4} x (x^{2} + y^{2})^{\frac{3}{2}} + \frac{3 y ^{2} x ( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}{8} + \frac{3 y ^{4}}{8} ln \frac{x + ( y ^{2} + x ^{2} ) ^{\frac{1}{2}}}{∣ y ∣} + C

n = 0 \sum \infty \frac{4 ^{n}}{4 ^{n}}

\frac{d}{d x} (lo g_{e} (x - 1))

\frac{d}{d x} (\frac{1}{4 + x})

\frac{d}{d x} ((4 x - 9) ln ((2 x - 5) (x - 2)))

\int_{2}^{1} (\frac{2}{x ^{2}} - 3 x^{2}) d x