What is the integral from 0 to 1 of (x+1)/(x^2+2x-3) ?

The integral from 0 to 1 of (x+1)/(x^2+2x-3) is diverges

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integral from 0 to 1 of (x+1)/(x^2+2x-3)

\int_{0}^{1} \frac{x + 1}{x ^{2} + 2 x - 3} d x

diverges

Solution steps

\int_{0}^{1} \frac{x + 1}{x ^{2} + 2 x - 3} d x

Apply u-substitution

: \int_{- 3}^{0} \frac{1}{2 u} d u

= \int_{- 3}^{0} \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_{- 3}^{0} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{- 3}^{0}

Compute the boundaries:

diverges

= diverges

\frac{d}{d x} (a x^{2} (x - 4) + b x^{4} ln (\frac{x}{4}))

\int x^{4} ln (x^{5}) d x

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

\frac{d}{d x} ((2 x - 1) sin (x - 2))

y = \frac{5}{ln ( x )}

What is the integral from 0 to 1 of (x+1)/(x^2+2x-3) ?
The integral from 0 to 1 of (x+1)/(x^2+2x-3) is diverges