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

The integral from 1 to 7 of (x+1)/(x^2+2x-1) is 1/2 ln(31)

Popular >

integral from 1 to 7 of (x+1)/(x^2+2x-1)

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

\frac{1}{2} ln (31)

Decimal

1.71699 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{62} \frac{1}{2 u} d u

= \int_{2}^{62} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{2}^{62} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{2}^{62}

Compute the boundaries:

ln (62) - ln (2)

= \frac{1}{2} (ln (62) - ln (2))

Simplify

= \frac{1}{2} ln (31)

y = \frac{π ^{2} x ^{6} + 9000}{x ^{2}}

\int 2 tan^{4} (x se) c^{6} x d x

f (x) = e^{x}, at x = 7

\frac{d}{d x} (\frac{3 x + 5}{1 + x})

\frac{d}{d x} (3^{x} \cdot ln (3))

What is the integral from 1 to 7 of (x+1)/(x^2+2x-1) ?
The integral from 1 to 7 of (x+1)/(x^2+2x-1) is 1/2 ln(31)