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

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

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

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

\frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

\int_{0}^{a r c t a n (x)} sec (u) d u = ln x + x^{2} + 1

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

Simplify

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

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

Compute the boundaries:

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

= \frac{1}{2} x x^{2} + 1 + \frac{1}{2} ln x + x^{2} + 1

Simplify

= \frac{x x ^{2} + 1 + ln x + x ^{2} + 1}{2}

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\int_{9}^{1} \frac{1}{x} d x

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