What is the integral of (14)/(x(x^2+25)) ?

The integral of (14)/(x(x^2+25)) is 14(1/25 ln|x|-1/50 ln|x^2+25|)+C

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integral of (14)/(x(x^2+25))

\int \frac{14}{x ( x ^{2} + 25 )} d x

14 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25) + C

Solution steps

\int \frac{14}{x ( x ^{2} + 25 )} d x

Take the constant out:

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

= 14 \cdot \int \frac{1}{x ( x ^{2} + 25 )} d x

Take the partial fraction of

\frac{1}{x ( x ^{2} + 25 )} : \frac{1}{25 x} - \frac{x}{25 ( x ^{2} + 25 )}

= 14 \cdot \int \frac{1}{25 x} - \frac{x}{25 ( x ^{2} + 25 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 14 (\int \frac{1}{25 x} d x - \int \frac{x}{25 ( x ^{2} + 25 )} d x)

\int \frac{1}{25 x} d x = \frac{1}{25} ln ∣ x ∣

\int \frac{x}{25 ( x ^{2} + 25 )} d x = \frac{1}{50} ln x^{2} + 25

= 14 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25)

Add a constant to the solution

= 14 (\frac{1}{25} ln ∣ x ∣ - \frac{1}{50} ln x^{2} + 25) + C

y^{^{''}} - 4 y^{^{'}} - 5 y = 0

x + y^{2} = 42, x = y

f (x) = x - 2, at x = 6

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 t ^{5}})

y = 4 cos (x), y = 6 - 2 cos (x), 0 \leq x \leq 2 π

What is the integral of (14)/(x(x^2+25)) ?
The integral of (14)/(x(x^2+25)) is 14(1/25 ln|x|-1/50 ln|x^2+25|)+C