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

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

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

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

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

Simplify

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

= 4 ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (\frac{x}{2})}^{0} - \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 x to 0 of sqrt(t^2+4) ?
The integral from x to 0 of sqrt(t^2+4) is 4(-(xsqrt(x^2+4))/8-1/2 ln((|x+sqrt(x^2+4)|)/2))