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

The integral from 1 to x^3 of 1/(1+t^2) is arctan(x^3)-pi/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to x^3 of 1/(1+t^2)

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

arctan (x^{3}) - \frac{π}{4}

Solution steps

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

Use the common integral:

\int \frac{1}{1 + t ^{2}} d t = arctan (t)

= [arctan (t)]_{1}^{x^{3}}

Compute the boundaries:

arctan (x^{3}) - \frac{π}{4}

= arctan (x^{3}) - \frac{π}{4}

\int_{0}^{t} e^{v} sin (t - v) d v

\int_{- 1}^{1} \frac{cos ( x )}{1 + x ^{4}} d x

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

\int_{0}^{π} 3 d x

\int_{1}^{0} ln (x) d x

What is the integral from 1 to x^3 of 1/(1+t^2) ?
The integral from 1 to x^3 of 1/(1+t^2) is arctan(x^3)-pi/4