What is the integral from 0 to 1 of 1/(sqrt(x^2+16)) ?

The integral from 0 to 1 of 1/(sqrt(x^2+16)) is ln(1/4+(sqrt(17))/4)

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integral from 0 to 1 of 1/(sqrt(x^2+16))

\int_{0}^{1} \frac{1}{x ^{2} + 16} d x

ln (\frac{1}{4} + \frac{17}{4})

Decimal Notation

0.24746 \dots

Solution steps

\int_{0}^{1} \frac{1}{x ^{2} + 16} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{1}{4})} sec (u) d u

= \int_{0}^{a r c t a n (\frac{1}{4})} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

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

Compute the boundaries:

ln (\frac{1}{4} + \frac{17}{4})

= ln (\frac{1}{4} + \frac{17}{4})

y^{^{''}} - 5 y^{^{'}} - 2 y = 0

\int_{- 8}^{8} (8 - ∣ x ∣) d x

h \to 0 lim ((x + h)^{2} - x^{2})

\int \frac{x + 1}{x ( x ^{2} + 1 )} d x

x \to - 6 lim (\frac{6 - ∣ x ∣}{6 + x})

What is the integral from 0 to 1 of 1/(sqrt(x^2+16)) ?
The integral from 0 to 1 of 1/(sqrt(x^2+16)) is ln(1/4+(sqrt(17))/4)