What is the integral from 0 to x^2 of y/(x^5+1) ?

The integral from 0 to x^2 of y/(x^5+1) is (x^4)/(2(1+x^5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to x^2 of y/(x^5+1)

\int_{0}^{x^{2}} \frac{y}{x ^{5} + 1} d y

\frac{x ^{4}}{2 ( 1 + x ^{5} )}

Solution steps

\int_{0}^{x^{2}} \frac{y}{x ^{5} + 1} d y

Take the constant out:

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

= \frac{1}{1 + x ^{5}} \cdot \int_{0}^{x^{2}} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{0}^{x^{2}}

= \frac{1}{1 + x ^{5}} [\frac{y ^{2}}{2}]_{0}^{x^{2}}

Simplify

\frac{1}{1 + x ^{5}} [\frac{y ^{2}}{2}]_{0}^{x^{2}} : \frac{[ \frac{y ^{2}}{2} ] _{0}^{x^{2}}}{1 + x ^{5}}

= \frac{[ \frac{y ^{2}}{2} ] _{0}^{x^{2}}}{1 + x ^{5}}

Compute the boundaries:

\frac{x ^{4}}{2}

= \frac{\frac{x ^{4}}{2}}{1 + x ^{5}}

Simplify

= \frac{x ^{4}}{2 ( 1 + x ^{5} )}

\int_{2}^{4} x^{2} + 2 x - 3 d x

\int_{1}^{3} \frac{1}{x ^{2} + x} d x

\int_{0}^{1} 6 x^{2} - 6 x^{3} d x

\int_{0}^{4} x^{2} e^{- x} d x

What is the integral from 0 to x^2 of y/(x^5+1) ?
The integral from 0 to x^2 of y/(x^5+1) is (x^4)/(2(1+x^5))