What is the integral from-x to x of sqrt(3+4t^2) ?

The integral from-x to x of sqrt(3+4t^2) is

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integral from-x to x of sqrt(3+4t^2)

\int_{- x}^{x} 3 + 4 t^{2} d t

x 3 + 4 x^{2} + \frac{3}{4} ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - \frac{3}{4} ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}

Solution steps

\int_{- x}^{x} 3 + 4 t^{2} d t

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

= \frac{3}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (- \frac{2 x}{3})}^{a r c t a n (\frac{2 x}{3})} + \frac{1}{2} (ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}))

Simplify

\frac{3}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (- \frac{2 x}{3})}^{a r c t a n (\frac{2 x}{3})} + \frac{1}{2} (ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3})) : \frac{3}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (- \frac{2 x}{3})}^{a r c t a n (\frac{2 x}{3})} + \frac{1}{2} (ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}))

= \frac{3}{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (- \frac{2 x}{3})}^{a r c t a n (\frac{2 x}{3})} + \frac{1}{2} (ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}))

Compute the boundaries:

2 \cdot \frac{x 3 + 4 x ^{2}}{3}

= \frac{3}{2} (2 \cdot \frac{x 3 + 4 x ^{2}}{3} + \frac{1}{2} (ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}))

Simplify

= x 3 + 4 x^{2} + \frac{3}{4} ln \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3} - \frac{3}{4} ln - \frac{2 x}{3} + \frac{3 + 4 x ^{2}}{3}

\int_{0}^{1} sinh (x) d x

\int_{\frac{1}{x}}^{4} 1 + \frac{1}{t} d t

\int_{2}^{4} e^{- s t} t^{2} d t

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

\int_{0}^{1} ln (r) d r

What is the integral from-x to x of sqrt(3+4t^2) ?
The integral from-x to x of sqrt(3+4t^2) is