What is the integral from 0 to 7 of 1/(sqrt(9+x^2)) ?

The integral from 0 to 7 of 1/(sqrt(9+x^2)) is ln(7/3+(sqrt(58))/3)

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integral from 0 to 7 of 1/(sqrt(9+x^2))

\int_{0}^{7} \frac{1}{9 + x ^{2}} d x

ln (\frac{7}{3} + \frac{58}{3})

Decimal Notation

1.58348 \dots

Solution steps

\int_{0}^{7} \frac{1}{9 + x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{a r c t a n (\frac{7}{3})} sec (u) d u

= \int_{0}^{a r c t a n (\frac{7}{3})} 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{7}{3})}

Compute the boundaries:

ln (\frac{7}{3} + \frac{58}{3})

= ln (\frac{7}{3} + \frac{58}{3})

\frac{d}{d x} (x^{2} - y)

\int \frac{x ^{2}}{( x ^{2} + 6 )} d x

y = (4 x^{2} + 5) (5 x - 3)

(3 x^{2} - 3 x + 1) (x^{3})

\int \frac{1}{( 4 + x ^{2} ) ^{\frac{5}{2}}} d x

What is the integral from 0 to 7 of 1/(sqrt(9+x^2)) ?
The integral from 0 to 7 of 1/(sqrt(9+x^2)) is ln(7/3+(sqrt(58))/3)