What is the integral of q/(pi(-0.125x+0.03125)^2) ?

The integral of q/(pi(-0.125x+0.03125)^2) is (8q)/(pi(0.03125-0.125x))+C

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integral of q/(pi(-0.125x+0.03125)^2)

\int \frac{q}{π ( - 0.125 x + 0.03125 ) ^{2}} d x

\frac{8 q}{π ( 0.03125 - 0.125 x )} + C

Solution steps

\int \frac{q}{π ( - 0.125 x + 0.03125 ) ^{2}} d x

Take the constant out:

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

= \frac{q}{π} \cdot \int \frac{1}{( 0.03125 - 0.125 x ) ^{2}} d x

Apply u-substitution

= \frac{q}{π} \cdot \int - \frac{1}{0.125 u ^{2}} d u

Take the constant out:

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

= \frac{q}{π} (- \frac{1}{0.125} \cdot \int \frac{1}{u ^{2}} d u)

Simplify

= \frac{q}{π} (- 8 \cdot \int \frac{1}{u ^{2}} d u)

Apply the Power Rule

= \frac{q}{π} (- 8 (- \frac{1}{u}))

Substitute back

u = 0.03125 - 0.125 x

= \frac{q}{π} (- 8 (- \frac{1}{0.03125 - 0.125 x}))

Simplify

\frac{q}{π} (- 8 (- \frac{1}{0.03125 - 0.125 x})) : \frac{8 q}{π ( 0.03125 - 0.125 x )}

= \frac{8 q}{π ( 0.03125 - 0.125 x )}

Add a constant to the solution

= \frac{8 q}{π ( 0.03125 - 0.125 x )} + C

What is the integral of q/(pi(-0.125x+0.03125)^2) ?
The integral of q/(pi(-0.125x+0.03125)^2) is (8q)/(pi(0.03125-0.125x))+C