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

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

\int_{0}^{2} 1 + x^{2} d x

\frac{3}{2} + \frac{1}{2} ln (2 + 3)

Solution steps

\int_{0}^{2} 1 + x^{2} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

= \int_{0}^{a r c t a n (2)} sec^{3} (u) d u

Apply Integral Reduction

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (2)} sec (u) d u

\int_{0}^{a r c t a n (2)} sec (u) d u = ln (2 + 3)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 3)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 3) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 3)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (2)} + \frac{1}{2} ln (2 + 3)

Compute the boundaries:

\frac{3}{2}

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

\int_{3}^{7} 4 x - 16 d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} tan^{2} (x) d x

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

\int_{0}^{1} (x + 4 1 - x^{2}) d x

\int_{- 1}^{1} 3 x^{3} d x

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