Indefinite Integrals
12. Integration of irrational algebraic functions
12. Integration of irrational algebraic functions
| S.No | Types of Integral | Substitution |
| 1. |
| Put $px + q = {t^2}$ |
| 2. | $$\int {\frac{{dx}}{{(ax + b)\sqrt {p{x^2} + qx + r} }}} $$ | Put $ax + b = \frac{1}{t}$ |
| 3. | $$\int {\frac{{dx}}{{(a{x^2} + b)\sqrt {p{x^2} + q} }}} $$ | Put $x = \frac{1}{t}$ |
| 4. |
| Put $x = \alpha {\cos ^2}\theta + \beta {\sin ^2}\theta $ |
| 5. |
| Put $x = \alpha {\sec ^2}\theta - \beta {\tan ^2}\theta $ |
| 6. | $$\int {\frac{{dx}}{{\sqrt {(x - \alpha )(x - \beta )} }}} $$ | Put $x - \alpha = {t^2}$ or $x - \beta = {t^2}$ |
| 7. | $$\int {\frac{{dx}}{{{{(x - k)}^r}\sqrt {a{x^2} + bx + c} }}} $$ where $r \geqslant 2,r \in I$ | Put $x - k = \frac{1}{t}$ |
| 8. | $$\int {\frac{{a{x^2} + bx + c}}{{(dx + e)\sqrt {f{x^2} + gx + h} }}dx} $$ | Write $a{x^2} + bx + c = {A_1}(dx + e)(2fx + g) + {B_1}(dx + e) + {C_1}$ where ${A_1},{B_1},{C_1}$ are constants and can be obtained by comparing the coefficients of like terms. |
Question 11. Evaluate $$\int {\frac{{x + 2}}{{({x^2} + 3x + 3)\sqrt {x + 1} }}dx} $$
Solution: $$I = \int {\frac{{x + 2}}{{({x^2} + 3x + 3)\sqrt {x + 1} }}dx} $$ Put $x + 1 = {t^2} \Rightarrow dx = 2tdt$. Therefore,
$$\begin{equation} \begin{aligned} I = \int {\frac{{\left( {{t^2} - 1} \right) + 2}}{{\left\{ {{{\left( {{t^2} - 1} \right)}^2} + 3\left( {{t^2} - 1} \right) + 3} \right\}\sqrt {{t^2}} }}dt} \\ I = 2\int {\frac{{{t^2} + 1}}{{{t^4} + {t^2} + 1}}} dt \\ I = 2\int {\frac{{1 + \frac{1}{{{t^2}}}}}{{{t^2} + 1 + \frac{1}{{{t^2}}}}}dt} \\ I = 2\int {\frac{{1 + \frac{1}{{{t^2}}}}}{{{{\left( {t - \frac{1}{t}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}}}dt} \\\end{aligned} \end{equation} $$ Put $t - \frac{1}{t} = u \Rightarrow \left( {1 + \frac{1}{{{t^2}}}} \right)dt = du$, we get
$$\begin{equation} \begin{aligned} I = 2\int {\frac{{du}}{{{{\left( u \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}}}} \\ I = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 3 }}} \right) + C \\ I = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{{t^2} - 1}}{{\sqrt 3 t}}} \right) + C \\ \therefore I = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt 3 (x + 1)}}} \right) + C \\\end{aligned} \end{equation} $$
