What is the integral of x^2sqrt(1+y^4) ?

The integral of x^2sqrt(1+y^4) is (x^3sqrt(1+y^4))/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2sqrt(1+y^4)

\int x^{2} 1 + y^{4} d x

\frac{x ^{3} 1 + y ^{4}}{3} + C

Solution steps

\int x^{2} 1 + y^{4} d x

Take the constant out:

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

= 1 + y^{4} \cdot \int x^{2} d x

Apply the Power Rule

: \frac{x ^{3}}{3}

= 1 + y^{4} \frac{x ^{3}}{3}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{x ^{3} 1 + y ^{4}}{3}

Add a constant to the solution

= \frac{x ^{3} 1 + y ^{4}}{3} + C

\int e^{x} (e^{x} + 1)^{3} d x

\int tan (5 x) sec (5 x) d x

\int \frac{x ^{3} + x + 1}{x ^{2} - 2 x + 1} d x

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

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

What is the integral of x^2sqrt(1+y^4) ?
The integral of x^2sqrt(1+y^4) is (x^3sqrt(1+y^4))/3+C