Question

Integrate $\int{\dfrac{u}{v}}dx$\[\]

Answer 1

Hint: We integrate the given integral using integration by parts $\left( \int{\left( f\left( x \right)g\left( x \right) \right)}dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( f{{\left( x \right)}^{'}}\int{g\left( x \right)dx} \right)}}dx \right)$ taking $\dfrac{1}{v}$ as the first function and $u$ as the second function. We differentiate $\dfrac{1}{v}$ using product rule of differentiation that is ${{\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}^{'}}=\dfrac{{{f}^{'}}\left( x \right)g\left( x \right)-f\left( x \right){{g}^{'}}\left( x \right)}{f\left( x \right)}$ and simplify. \[\]

Complete step-by-step answer:
We know that if there two single variable real valued integrable functions in product form say $f$ and $g$ then we integrate them by parts taking $f$ as first function and $g$as second function using the formula
\[\int{\left( f\left( x \right)g\left( x \right) \right)}dx=f\left( x \right)\int{g\left( x \right)dx-\int{\left( f{{\left( x \right)}^{'}}\int{g\left( x \right)dx} \right)}}dx\]
The choice of the first function depends upon how many times differentiating the function will make zero. So the rule that is used when we are integrating by parts is called ILATE, an acronym for inverse, algebraic, logarithm, trigonometric and finally exponent. It means we have to choose the first function in the order of ILATE. \[\]

If we have to differentiate product of two functions say $f\left( x \right),g\left( x \right)$ which are in quotient $\dfrac{f\left( x \right)}{g\left( x \right)}$ form , we use the quotient rule of differentiation

\[{{\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}^{'}}=\dfrac{{{f}^{'}}\left( x \right)g\left( x \right)-f\left( x \right){{g}^{'}}\left( x \right)}{f\left( x \right)}\]

The given integral is $\int{\dfrac{u}{v}}dx$ and the integrand is $\dfrac{u}{v}$. Where we assume that $u$ and $v$ are integrable functions and $v\ne 0$. We shall use the product form formula of the given integral taking the first function as $\dfrac{1}{v}$ and the second function as $u$ . Now we integrate by parts,
\[\int{\left( \dfrac{1}{v}\cdot u \right)}dx=\dfrac{1}{v}\int{udx-\int{\left( \dfrac{d}{dx}\left( \dfrac{1}{v} \right)\int{udx} \right)}}dx\]
Let us substitute $U=\int{udx}$ in above equation,
\[\int{\left( \dfrac{1}{v}\cdot u \right)}dx=\dfrac{1}{v}U-\left( \dfrac{d}{dx}\left( \dfrac{1}{v} \right)U \right)...(1)\]
We differentiate the function $\dfrac{1}{v}$ using the quotient rule of differentiation to get,
\[\dfrac{d}{dx}\left( \dfrac{1}{v} \right)=\dfrac{v.0-1.{{v}^{'}}}{{{v}^{2}}}=\dfrac{-{{v}^{'}}}{{{v}^{2}}}\]
 Putting it in above equation(1) we get,
\[\begin{align}
  & \int{\left( \dfrac{1}{v}\cdot u \right)}dx=\dfrac{1}{v}U-\left( \dfrac{-{{v}^{'}}}{{{v}^{2}}}U \right). \\
 & =\left( \dfrac{1}{v}+\dfrac{{{v}^{'}}}{{{v}^{2}}} \right)U \\
 & =\left( \dfrac{1}{v}+\dfrac{{{v}^{'}}}{{{v}^{2}}} \right)\int{udx} \\
\end{align}\]
So $\int{\dfrac{u}{v}}dx==\left( \dfrac{1}{v}+\dfrac{{{v}^{'}}}{{{v}^{2}}} \right)\int{udx}$.\[\]

Note: The given integral is called quotient rule of integration. The functions in integrand need to be continuous and differentiable otherwise the integration by parts cannot be used. When three functions are in product form we can integrate them using the expression $uvw=\int{uv{{w}^{'}}}+\int{u{{v}^{'}}w}+\int{{{u}^{'}}vw}$.