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

The integral from 0 to 3 of 1/(xsqrt(x+1)) is diverges

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integral from 0 to 3 of 1/(xsqrt(x+1))

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

diverges

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

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

Simplify

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

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

Compute the boundaries:

diverges

= diverges

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