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

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

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

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

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Simplify

= \frac{π}{6}

x \to 0 + lim (x^{x^{2}})

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

\int e^{15 x} (15) d x

x \to 4 lim (\frac{x - 4}{x - 2})

f (x) = cos (x), g (x) = - cos (x) + 2

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