What is the integral from 0 to x of sqrt(t^2+4) ?

The integral from 0 to x of sqrt(t^2+4) is 4((xsqrt(x^2+4))/8+1/2 ln((|x+sqrt(x^2+4)|)/2))

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integral from 0 to x of sqrt(t^2+4)

\int_{0}^{x} t^{2} + 4 d t

4 (\frac{x x ^{2} + 4}{8} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

Solution steps

\int_{0}^{x} t^{2} + 4 d t

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{x}{2})} 4 sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{x}{2})} 4 sec^{3} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{a r c t a n (\frac{x}{2})} sec^{3} (u) d u

Apply Integral Reduction

= 4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{x}{2})} sec (u) d u)

\int_{0}^{a r c t a n (\frac{x}{2})} sec (u) d u = ln (\frac{x + x ^{2} + 4}{2})

= 4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

Simplify

4 ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2})) : 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

= 4 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{x}{2})} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

Compute the boundaries:

\frac{x x ^{2} + 4}{8}

= 4 (\frac{x x ^{2} + 4}{8} + \frac{1}{2} ln (\frac{x + x ^{2} + 4}{2}))

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What is the integral from 0 to x of sqrt(t^2+4) ?
The integral from 0 to x of sqrt(t^2+4) is 4((xsqrt(x^2+4))/8+1/2 ln((|x+sqrt(x^2+4)|)/2))