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

The integral from 1 to 3 of 1/(sqrt(x)(x+1)) is pi/6

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integral from 1 to 3 of 1/(sqrt(x)(x+1))

\int_{1}^{3} \frac{1}{x ( x + 1 )} d x

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

\int_{1}^{3} \frac{1}{x ( x + 1 )} d x

Apply u-substitution

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

= \int_{1}^{3} \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}^{3} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 2 [arctan (u)]_{1}^{3}

Compute the boundaries:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Simplify

= \frac{π}{6}

\int_{1}^{x^{2}} sin^{2} (t) d t

\frac{\partial}{\partial x} ((x + 3 y)^{3})

(a + \frac{b}{x ^{2}})^{3}

f (x) = e^{4 x} + 4 e^{4 x} x

x \to \infty lim (\frac{ln ^{3} ( x )}{100 3 x})

What is the integral from 1 to 3 of 1/(sqrt(x)(x+1)) ?
The integral from 1 to 3 of 1/(sqrt(x)(x+1)) is pi/6