What is the integral of+(2x+1)/((x+5)^3) ?

The integral of+(2x+1)/((x+5)^3) is -2/(x+5)+9/(2(x+5)^2)+C

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integral of+(2x+1)/((x+5)^3)

\int + \frac{2 x + 1}{( x + 5 ) ^{3}} d x

- \frac{2}{x + 5} + \frac{9}{2 ( x + 5 ) ^{2}} + C

Solution steps

\int \frac{2 x + 1}{( x + 5 ) ^{3}} d x

Apply u-substitution

: \int \frac{2 u - 9}{u ^{3}} d u

= \int \frac{2 u - 9}{u ^{3}} d u

Expand

\frac{2 u - 9}{u ^{3}} : \frac{2}{u ^{2}} - \frac{9}{u ^{3}}

= \int \frac{2}{u ^{2}} - \frac{9}{u ^{3}} d u

Apply the Sum Rule:

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

= \int \frac{2}{u ^{2}} d u - \int \frac{9}{u ^{3}} d u

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

\int \frac{9}{u ^{3}} d u = - \frac{9}{2 u ^{2}}

= - \frac{2}{u} - (- \frac{9}{2 u ^{2}})

Substitute back

u = x + 5

= - \frac{2}{x + 5} - (- \frac{9}{2 ( x + 5 ) ^{2}})

Simplify

= - \frac{2}{x + 5} + \frac{9}{2 ( x + 5 ) ^{2}}

Add a constant to the solution

= - \frac{2}{x + 5} + \frac{9}{2 ( x + 5 ) ^{2}} + C

x \cdot \frac{d y}{d x} + 5 y = 7 x^{2}

\frac{d}{d y} (x - y^{3} + y^{2} sin (x))

x^{3} + 5

y \to - 3 lim ((5 - y)^{\frac{4}{3}})

What is the integral of+(2x+1)/((x+5)^3) ?
The integral of+(2x+1)/((x+5)^3) is -2/(x+5)+9/(2(x+5)^2)+C