What is the integral from 0 to 1 of x/(x^2+9) ?

The integral from 0 to 1 of x/(x^2+9) is 1/2 ln(10)-ln(3)

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integral from 0 to 1 of x/(x^2+9)

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

\frac{1}{2} ln (10) - ln (3)

Decimal Notation

0.05268 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} + 9} d x

Apply u-substitution

: \int_{9}^{10} \frac{1}{2 u} d u

= \int_{9}^{10} \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{9}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{9}^{10}

Compute the boundaries:

ln (10) - 2 ln (3)

= \frac{1}{2} (ln (10) - 2 ln (3))

Simplify

= \frac{1}{2} ln (10) - ln (3)

\int \frac{1}{( 3 x + 1 ) ^{2}} d x

e^{(x - 1)^{2}}

\int_{0}^{\infty} t^{2} d t

\int \frac{x ^{- \frac{1}{2}}}{1 + x ^{\frac{1}{2}}} d x

\frac{1}{4 x - 3}

What is the integral from 0 to 1 of x/(x^2+9) ?
The integral from 0 to 1 of x/(x^2+9) is 1/2 ln(10)-ln(3)