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

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

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

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

diverges

Solution steps

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

Complete the square

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

= \int_{0}^{1} \frac{1}{( x + 1 ) ^{2} - 1} d x

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{1}^{2}

Simplify

- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{1}^{2} : - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{1}^{2}

= - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{1}^{2}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{24}} x tan^{2} (8 x) d x

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

\int_{- 1}^{125} \frac{1}{x ^{\frac{1}{3}}} d x

\int_{2}^{3} 10 x d x

\int_{- 2}^{5} x^{2} + 2 x + 4 d x

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