What is the integral from 0 to x^2 of 1/(3+x^3) ?

The integral from 0 to x^2 of 1/(3+x^3) is (x^2)/(x^3+3)

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 1/(3+x^3)

\int_{0}^{x^{2}} \frac{1}{3 + x ^{3}} d y

\frac{x ^{2}}{x ^{3} + 3}

Solution steps

\int_{0}^{x^{2}} \frac{1}{3 + x ^{3}} d y

Integral of a constant:

\int a d x = a x

= [\frac{1}{3 + x ^{3}} y]_{0}^{x^{2}}

Multiply

\frac{1}{3 + x ^{3}} y : \frac{y}{3 + x ^{3}}

= [\frac{y}{3 + x ^{3}}]_{0}^{x^{2}}

Compute the boundaries:

\frac{x ^{2}}{x ^{3} + 3}

= \frac{x ^{2}}{x ^{3} + 3}

\int_{0}^{7} 8 d x

\int_{\frac{π}{2}}^{π} x cos (x) d x

\int_{1}^{3} \frac{1}{y} d y

\int_{0}^{1} x arctan^{2} (x) d x

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

What is the integral from 0 to x^2 of 1/(3+x^3) ?
The integral from 0 to x^2 of 1/(3+x^3) is (x^2)/(x^3+3)