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

The integral from 1 to 3 of sqrt(1+t^2) is (sqrt(2)(-1+3sqrt(5))-ln(1+sqrt(2))+ln(3+sqrt(10)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

Solution steps

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

Apply Trigonometric Substitution

: \int_{\frac{π}{4}}^{a r c t a n (3)} sec^{3} (u) d u

= \int_{\frac{π}{4}}^{a r c t a n (3)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} \cdot \int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u

\int_{\frac{π}{4}}^{a r c t a n (3)} sec (u) d u = ln (3 + 10) - ln (1 + 2)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2)) : [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

= [\frac{1}{2} sec (u) tan (u)]_{\frac{π}{4}}^{a r c t a n (3)} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Compute the boundaries:

\frac{3 5 - 1}{2}

= \frac{3 5 - 1}{2} + \frac{1}{2} (ln (3 + 10) - ln (1 + 2))

Simplify

= \frac{2 ( - 1 + 3 5 ) - ln ( 1 + 2 ) + ln ( 3 + 10 )}{2}

\int_{0}^{1} \frac{2}{2 - r ^{2}} r d r

\int_{- 2}^{2} (x^{3} - x^{5}) d x

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

\int_{0}^{4} (y - 2)^{2} d y

\int_{2}^{4} \frac{1}{x ^{2} - 3} d x

What is the integral from 1 to 3 of sqrt(1+t^2) ?
The integral from 1 to 3 of sqrt(1+t^2) is (sqrt(2)(-1+3sqrt(5))-ln(1+sqrt(2))+ln(3+sqrt(10)))/2