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

The integral from-1 to 7 of 1/((x+1)) is diverges

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{8} \frac{1}{u} d u

= \int_{0}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{8}

Compute the boundaries:

diverges

= diverges

\int_{- 5}^{6} (1 - x) d x

\int_{- 1}^{2} x^{3} d x

\int_{0}^{4} - x + 4 d x

\int_{0}^{1} x^{2} e^{x y} d y

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

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