What is the integral of 4/(x^2-14x+49) ?

The integral of 4/(x^2-14x+49) is -4/(x-7)+C

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integral of 4/(x^2-14x+49)

\int \frac{4}{x ^{2} - 14 x + 49} d x

- \frac{4}{x - 7} + C

Solution steps

\int \frac{4}{x ^{2} - 14 x + 49} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} - 14 x + 49} d x

Take the partial fraction of

\frac{1}{x ^{2} - 14 x + 49} : \frac{1}{( x - 7 ) ^{2}}

= 4 \cdot \int \frac{1}{( x - 7 ) ^{2}} d x

Apply u-substitution

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

= 4 \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 4 (- \frac{1}{u})

Substitute back

u = x - 7

= 4 (- \frac{1}{x - 7})

Simplify

4 (- \frac{1}{x - 7}) : - \frac{4}{x - 7}

= - \frac{4}{x - 7}

Add a constant to the solution

= - \frac{4}{x - 7} + C

\frac{\partial}{\partial x} (cos^{2} (5 x))

\frac{d}{d x} (x^{- \frac{3}{7}})

5 y^{^{''}} + y^{^{'}} = - 6 x, y (0) = 0, y^{^{'}} (0) = - 20

f (x) = 6 tan (x), at x = \frac{π}{3}

\int sin^{3} (2 x) cos^{8} (2 x) d x

What is the integral of 4/(x^2-14x+49) ?
The integral of 4/(x^2-14x+49) is -4/(x-7)+C