What is the integral from 0 to 3 of sqrt(x^2+16) ?

The integral from 0 to 3 of sqrt(x^2+16) is 16(15/32+1/2 ln(2))

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integral from 0 to 3 of sqrt(x^2+16)

\int_{0}^{3} x^{2} + 16 d x

16 (\frac{15}{32} + \frac{1}{2} ln (2))

Decimal Notation

13.04517 \dots

Solution steps

\int_{0}^{3} x^{2} + 16 d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{3}{4})} 16 sec^{3} (u) d u

= \int_{0}^{a r c t a n (\frac{3}{4})} 16 sec^{3} (u) d u

Take the constant out:

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

= 16 \cdot \int_{0}^{a r c t a n (\frac{3}{4})} sec^{3} (u) d u

Apply Integral Reduction

= 16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (\frac{3}{4})} sec (u) d u

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

= 16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2)

Simplify

16 [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2) : 16 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2))

= 16 ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (\frac{3}{4})} + \frac{1}{2} ln (2))

Compute the boundaries:

\frac{15}{32}

= 16 (\frac{15}{32} + \frac{1}{2} ln (2))

\frac{d y}{d x} + \frac{y}{x} = \frac{- y ^{2}}{x}

\int \frac{4 x}{( 2 x ^{2} + 3 ) ^{3}} d x

\int \frac{x ^{2} - 3}{x + 1} d x

\int \frac{1}{4 x ^{2} + 1} d x

x \to 1 lim (3 f (x) - 2 h (x))

What is the integral from 0 to 3 of sqrt(x^2+16) ?
The integral from 0 to 3 of sqrt(x^2+16) is 16(15/32+1/2 ln(2))