What is the integral of 1/((a-x)^2) ?

The integral of 1/((a-x)^2) is 1/(a-x)+C

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integral of 1/((a-x)^2)

\int \frac{1}{( a - x ) ^{2}} d x

\frac{1}{a - x} + C

Solution steps

\int \frac{1}{( a - x ) ^{2}} d x

Apply u-substitution

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

= \int - \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 \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - (- \frac{1}{u})

Substitute back

u = a - x

= - (- \frac{1}{a - x})

Simplify

= \frac{1}{a - x}

Add a constant to the solution

= \frac{1}{a - x} + C

\frac{d y}{d x}, 2 x y - 4 y = x y - x y^{2}

y^{^{'}} = \frac{9 x ^{2} - 6}{x ^{2}}

5 x - x^{2}, x, 0, 4

n = 3 \sum \infty \frac{5 ^{n}}{1 4 ^{n}}

\int x \cdot e^{4 x} d x