Indefinite Integrals
6.0 Integral of the type $\int {\frac{{px + q}}{{a{x^2} + bx + c}}dx,} \int {\frac{{px + q}}{{\sqrt {a{x^2} + bx + c} }}dx,} \int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
6.0 Integral of the type $\int {\frac{{px + q}}{{a{x^2} + bx + c}}dx,} \int {\frac{{px + q}}{{\sqrt {a{x^2} + bx + c} }}dx,} \int {(px + q)} \sqrt {a{x^2} + bx + c} dx$
In this type of integral, the linear factor $px+q$ is written in terms of the derivative of ${a{x^2} + bx + c}$ i.e., $$\begin{equation} \begin{aligned} px + q = A\frac{d}{{dx}}\left( {a{x^2} + bx + c} \right) + B \\ \Rightarrow px + q = A\left( {2ax + b} \right) + B\quad ...(1) \\\end{aligned} \end{equation} $$ where $A$ and $B$ are constant and their values are found out using equation $(1)$.
Note: If the given integral is of the type $$\int {\frac{{f(x)}}{{a{x^2} + bx + c}}dx} $$ where $f(x)$ is a polynomial of degree greater than $2$, then we divide the numerator by denominator to convert the given integral in the form $$g(x) + \frac{{h(x)}}{{a{x^2} + bx + c}}$$ such that $h(x)$ becomes a linear function of $x$ and the above mentioned technique to solve the integral with linear factor $px+q$ is applied.
Question 5. Evaluate $$\int {(x - 5)\sqrt {{x^2} + x} } dx$$
Solution: Let us assume $$x - 5 = A\frac{d}{{dx}}\left( {{x^2} + x} \right) + B$$ Then, $$x - 5 = A(2x + 1) + B$$
On comparing the coefficients, we get
$1=2A$ and $A+B=-5$
$ \Rightarrow A = \frac{1}{2},B = - \frac{{11}}{2}$
Therefore,
$$\begin{equation} \begin{aligned} I = \int {(x - 5)\sqrt {{x^2} + x} } dx = \int {\left( {\frac{1}{2}(2x + 1) - \frac{{11}}{2}} \right)\sqrt {{x^2} + x} } dx \\ I = \int {\left( {\frac{1}{2}(2x + 1)} \right)\sqrt {{x^2} + x} } dx - \frac{{11}}{2}\int {\sqrt {{x^2} + x} } dx \\ I = {I_1} + {I_2} \\\end{aligned} \end{equation} $$
Take first integral, $${I_1} = \int {\left( {\frac{1}{2}(2x + 1)} \right)\sqrt {{x^2} + x} } dx$$
Put ${x^2} + x = t \Rightarrow (2x + 1)dx = dt$ $$\begin{equation} \begin{aligned} \therefore {I_1} = \frac{1}{2}\int {\sqrt t } dt = \frac{1}{2}\frac{{{{\left( t \right)}^{\frac{3}{2}}}}}{{3/2}} + {C_1} \\ {I_1} = \frac{1}{3}{\left( {{x^2} + x} \right)^{\frac{3}{2}}} + {C_1} \\\end{aligned} \end{equation} $$
Take second integral, we get
$$ {I_2} = - \frac{{11}}{2}\int {\sqrt {{x^2} + x} } dx $$ $$ {I_2} = - \frac{{11}}{2}\int {\sqrt {{{\left( {x + \frac{1}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } dx $$ $$ {I_2} = - \frac{{11}}{2}\left[ {\left\{ {\frac{1}{2}\left( {x + \frac{1}{2}} \right)\sqrt {{{\left( {x + \frac{1}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } \right\} - \frac{1}{2}.{{\left( {\frac{1}{2}} \right)}^2}\ln \left[ {\left( {x + \frac{1}{2}} \right) + \sqrt {{{\left( {x + \frac{1}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } \right]} \right] + {C_2} $$ $$ {I_2} = - \frac{{11}}{2}\left[ {\frac{{2x + 1}}{4}\sqrt {{x^2} + x} - \frac{1}{8}\ln \left| {\left( {x + \frac{1}{2}} \right) + \sqrt {{x^2} + x} } \right|} \right] + {C_2} $$
Therefore,
$$I = {I_1} + {I_2} = \frac{1}{3}{\left( {{x^2} + x} \right)^{\frac{3}{2}}} - \frac{{11}}{2}\left[ {\frac{{2x + 1}}{4}\sqrt {{x^2} + x} - \frac{1}{8}\ln \left| {\left( {x + \frac{1}{2}} \right) + \sqrt {{x^2} + x} } \right|} \right] + C$$
