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

The integral from 0 to 3 of sqrt(x^2+1) is (3sqrt(5))/(sqrt(2))+1/2 ln(3+sqrt(10))

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integral from 0 to 3 of sqrt(x^2+1)

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

\int_{0}^{a r c t a n (3)} sec (u) d u = ln (3 + 10)

= [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

Simplify

[\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10) : [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

= [\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (3)} + \frac{1}{2} ln (3 + 10)

Compute the boundaries:

\frac{3 5}{2}

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

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

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

\int_{- 3}^{0} x + 3 + 9 - x^{2} d x

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

\int_{0}^{\frac{π}{2}} (cos (2 x)) d x

What is the integral from 0 to 3 of sqrt(x^2+1) ?
The integral from 0 to 3 of sqrt(x^2+1) is (3sqrt(5))/(sqrt(2))+1/2 ln(3+sqrt(10))