The integrals of irrational functions are rarely represented by well-known functions. We can calculate only if the functions involve special forms such that \(\sqrt[n]{ax+b}, \sqrt[n]{\frac{ax+b}{cx+d}}\) and \(\sqrt{ax^{2}+bx+c}\).
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Forms Involving \(\sqrt[n]{ax+b}\)
In the case of \(\int f(x,\sqrt[n]{ax+b})\ dx\), substitute \(t=\sqrt[n]{ax+b}\).
Example: Find \(\int \frac{1}{x\sqrt{x+1}}\ dx\)
Solution : Let \(\sqrt{x+1}=t\). Then we have \(x=t^{2}-1\) and \(dx=2t\ dt\).
Thus, the Substitution Rule gives
\begin{eqnarray*}&&\ \int \frac{1}{x\sqrt{x+1}}\ dx=\int \frac{1}{(t^{2}-1)t}2t\ dt=\int \frac{2}{t^{2}-1}\ dt=\int \left(\frac{1}{t-1}-\frac{1}{t+1}\right)\ dt\\&&=\ln |t-1|-\ln |t+1|+C=\ln\left|\frac{t-1}{t+1}\right|+C=\ln\left|\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right|+C\end{eqnarray*}
Forms Involving \(\sqrt[n]{\frac{ax+b}{cx+d}}\)
In the case of \(\int f(x,\sqrt[n]{\frac{ax+b}{cx+d}})\ dx\), substitute \(t=\sqrt[n]{\frac{ax+b}{cx+d}}\).
Example: Find \(\int \frac{1}{x}\sqrt{\frac{x+1}{x-1}}\ dx\)
Solution : Let \(\sqrt{\frac{x+1}{x-1}}=t\). Then we have \(x=\frac{t^{2}+1}{t^{2}-1}\) and \(dx=-\frac{4t}{(t^{2}-1)^{2}}\ dt\).
Thus, the Substitution Rule gives
\begin{eqnarray*}&&\ \int \frac{1}{x}\sqrt{\frac{x+1}{x-1}}\ dx=\int \frac{t^{2}-1}{t^{2}+1}\cdot t \cdot \frac{-4t}{(t^{2}-1)^{2}}\ dt=\int \frac{-4t^{2}}{(t^{2}+1)(t^{2}-1)}\ dt\\&&\\&& =\int \left(-\frac{2}{t^{2}+1}+\frac{1}{t+1}-\frac{1}{t-1}\right)\ dt=-2\tan^{-1} t+\ln \left|\frac{t+1}{t-1}\right|+C\\&&\\ &&=-2\tan^{-1} \sqrt{\frac{x+1}{x-1}}+\ln \left|\frac{\sqrt{\frac{x+1}{x-1}}+1}{\sqrt{\frac{x+1}{x-1}}-1}\right|+C=-2\tan^{-1}\sqrt{\frac{x+1}{x-1}}+2\ln(\sqrt{x+1}+\sqrt{x-1})+C\end{eqnarray*}
Forms Involving \(\sqrt{ax^{2}+bx+c}\)
In the case of \(a>0 \), substitute \(\sqrt{ax^{2}+bx+c}=t-\sqrt{a}x\).
Example: Find \(\int \frac{1}{x\sqrt{x^{2}+x+1}}\ dx\)
Solution : Let \(\sqrt{x^{2}+x+1}=t-x\). Then we have
$$x=\frac{t^{2}-1}{2t+1},\ \ \ \sqrt{x^{2}+x+1}=\frac{t^{2}+t+1}{2t+1},\ \ \ dx=\frac{2t^{2}+2t+2}{(2t+1)^{2}}dt$$
Thus, the Substitution Rule gives
$$\int \frac{1}{x\sqrt{x^{2}+x+1}}\ dx=\int \frac{2}{t^{2}-1}\ dt=\ln\left|\frac{t-1}{t+1}\right|+C=\ln\left|\frac{\sqrt{x^{2}+x+1}+x-1}{\sqrt{x^{2}+x+1}+x+1}\right|+C$$
In the case of \(a\) < \(0 \) involving \(\sqrt{ax^{2}+bx+c}=a(x-\alpha)(x-\beta)\) , substitute \(\sqrt{\frac{a(x-\alpha)}{x-\beta}}=t\) (Suppose \(\alpha\) < \(\beta\)).
Example: Find \(\int \frac{x}{\sqrt{(5-4x-x^{2})^{3}}}\ dx\)
Solution : Since \((5-4x-x^{2})=-(x+5)(x-1)\), we let \(t=\sqrt{\frac{-(x+5)}{x-1}}\) where \(a=-1,\ \alpha=-5,\ \beta=1\). Then we have
$$x=\frac{t^{2}-5}{t^{2}+1},\ \ \ \sqrt{5-4x-x^{2}}=(1-x)t=\frac{6t}{t^{2}+1},\ \ \ dx=\frac{12t}{(t^{2}+1)^{2}}dt$$
Thus, the Substitution Rule gives
\begin{eqnarray*}\int \frac{x}{\sqrt{(5-4x-x^{2})^{3}}}\ dx&=&\int \frac{\ \ \ \frac{t^{2}-5}{t^{2}+1}\ \ \ }{\ \ \ \frac{216t^{3}}{(t^{2}+1)^{3}}\ \ \ }\frac{12t}{(t^{2}+1)^{2}}dt=\frac{1}{18}\int \left(1-\frac{5}{t^{2}}\right)\ dt\\ && \\&=&\frac{1}{18}\left(t+\frac{5}{t}\right)+C=\frac{5-2x}{9\sqrt{5-4x-x^{2}}}+C\end{eqnarray*}