What is the integral from 1 to 6 of 1/(sqrt(x^2+2x)) ?

The integral from 1 to 6 of 1/(sqrt(x^2+2x)) is ln(4sqrt(3)+7)-ln(sqrt(3)+2)

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integral from 1 to 6 of 1/(sqrt(x^2+2x))

\int_{1}^{6} \frac{1}{x ^{2} + 2 x} d x

ln (43 + 7) - ln (3 + 2)

Decimal

1.31695 \dots

Solution steps

\int_{1}^{6} \frac{1}{x ^{2} + 2 x} d x

Complete the square

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

= \int_{1}^{6} \frac{1}{( x + 1 ) ^{2} - 1} d x

Apply u-substitution

: \int_{2}^{7} \frac{1}{u ^{2} - 1} d u

= \int_{2}^{7} \frac{1}{u ^{2} - 1} d u

Apply Trigonometric Substitution

: \int_{\frac{π}{3}}^{arcsec (7)} sec (v) d v

= \int_{\frac{π}{3}}^{arcsec (7)} sec (v) d v

Use the common integral:

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

= [ln ∣ tan (v) + sec (v) ∣]_{\frac{π}{3}}^{arcsec (7)}

Compute the boundaries:

ln (43 + 7) - ln (3 + 2)

= ln (43 + 7) - ln (3 + 2)

\int_{1}^{\infty} \frac{1}{x - 1} d x

\int_{0}^{1} \frac{1}{4 - x} d x

\int_{6}^{11} 4 x - 10 x e^{- 0.2 x} d x

\int_{1}^{3} 5 x + 4 d x

\int_{- \infty}^{0} \frac{1}{x ^{2} + 4} d x

What is the integral from 1 to 6 of 1/(sqrt(x^2+2x)) ?
The integral from 1 to 6 of 1/(sqrt(x^2+2x)) is ln(4sqrt(3)+7)-ln(sqrt(3)+2)