What is the integral of x/(10-x^2) ?

The integral of x/(10-x^2) is -1/2 ln|10-x^2|+C

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integral of x/(10-x^2)

\int \frac{x}{10 - x ^{2}} d x

- \frac{1}{2} ln 10 - x^{2} + C

Solution steps

\int \frac{x}{10 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 10 - x^{2}

= - \frac{1}{2} ln 10 - x^{2}

Add a constant to the solution

= - \frac{1}{2} ln 10 - x^{2} + C

n = 1 \sum \infty \frac{1}{( 2 n - 1 )}

\int 7 sin^{4} (x) cos^{2} (x) d x

\frac{\partial}{\partial x} (63 x^{8} y^{6} - 48 x^{7} y^{5})

f (x) = (x^{2} + 1)^{4}

y^{^{'}} = 0.15 ln (\frac{2000}{y}) y

What is the integral of x/(10-x^2) ?
The integral of x/(10-x^2) is -1/2 ln|10-x^2|+C