What is the integral of 9/(x^2(x^2+25)) ?

The integral of 9/(x^2(x^2+25)) is 9(-1/(25x)-1/125 arctan(x/5))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9/(x^2(x^2+25))

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

Solution steps

\int \frac{9}{x ^{2} ( x ^{2} + 25 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ^{2} ( x ^{2} + 25 )} d x

Take the partial fraction of

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

= 9 \cdot \int \frac{1}{25 x ^{2}} - \frac{1}{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

= 9 (\int \frac{1}{25 x ^{2}} d x - \int \frac{1}{25 ( x ^{2} + 25 )} d x)

\int \frac{1}{25 x ^{2}} d x = - \frac{1}{25 x}

\int \frac{1}{25 ( x ^{2} + 25 )} d x = \frac{1}{125} arctan (\frac{x}{5})

= 9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5}))

Add a constant to the solution

= 9 (- \frac{1}{25 x} - \frac{1}{125} arctan (\frac{x}{5})) + C

\frac{1}{x}

\frac{d x}{d t} = 6 - \frac{3 x}{500}

\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

\frac{\partial}{\partial x} (x^{\frac{2}{3}})

x \to 0 + lim (e^{x} + ln (x))

What is the integral of 9/(x^2(x^2+25)) ?
The integral of 9/(x^2(x^2+25)) is 9(-1/(25x)-1/125 arctan(x/5))+C