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

The integral from 0 to a of 1/((a+r-x)^2) is a/(r(a+r))

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integral from 0 to a of 1/((a+r-x)^2)

\int_{0}^{a} \frac{1}{( a + r - x ) ^{2}} d x

\frac{a}{r ( a + r )}

Solution steps

\int_{0}^{a} \frac{1}{( a + r - x ) ^{2}} d x

Apply u-substitution

: \int_{a + r}^{r} - \frac{1}{u ^{2}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: [- \frac{1}{u}]_{a + r}^{r}

= - [- \frac{1}{u}]_{a + r}^{r}

Compute the boundaries:

- \frac{1}{r} + \frac{1}{a + r}

= - (- \frac{1}{r} + \frac{1}{a + r})

Simplify

= \frac{a}{r ( a + r )}

x, x^{3}

\int \frac{sin ( x )}{cos ^{6} ( x )} d x

- \frac{20}{s ( s + 10 )}

y = 2 x^{3} + 3 x^{2} - 12 x + 3

\frac{d}{d x} (5 x^{6} - 12)

What is the integral from 0 to a of 1/((a+r-x)^2) ?
The integral from 0 to a of 1/((a+r-x)^2) is a/(r(a+r))