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

The integral of ((1+v^2))/(v^3) is -1/(2v^2)+ln|v|+C

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integral of ((1+v^2))/(v^3)

integral

\frac{( 1 + v ^{2} )}{v ^{3}}

- \frac{1}{2 v ^{2}} + ln ∣ v ∣ + C

Solution steps

\int \frac{1 + v ^{2}}{v ^{3}} d v

Expand

\frac{1 + v ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{1}{v}

= \int \frac{1}{v ^{3}} + \frac{1}{v} d v

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{1}{v ^{3}} d v + \int \frac{1}{v} d v

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

\int \frac{1}{v} d v = ln ∣ v ∣

= - \frac{1}{2 v ^{2}} + ln ∣ v ∣

Add a constant to the solution

= - \frac{1}{2 v ^{2}} + ln ∣ v ∣ + C

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

\frac{5}{( 3 + 4 x ^{2} )}

\frac{1}{( x ^{\frac{1}{3}} + 1 )}

\frac{1}{( y ^{3} - 3 y )}

- 1^{x}

What is the integral of ((1+v^2))/(v^3) ?
The integral of ((1+v^2))/(v^3) is -1/(2v^2)+ln|v|+C