Integration of irrational functions

Integrals of the form $$\int R\left(x, \left(\frac{ax+b}{cx+d}\right)^{\frac{m_1}{n_1}},\,\left(\frac{ax+b}{cx+d}\right)^{\frac{m_2}{n_2}},\,...\right)dx$$ where $R(x, y, z, ...)$ is a rational function of its arguments, $m_1,, n_1, , m_2,, n_2,... -$ are integers, are evaluated using the substitution $\frac{ax+b}{cx+d}=t^s,$ where $s -$ is the least common denominator of the fractions $\frac{m_1}{n_1},,,\frac{m_2}{n_2},,, ...$

The computation of integrals of the form $$\int R(x,\sqrt{ax^2+bx+c})\,dx,$$ where $R -$ is a rational function of two arguments, is carried out using trigonometric substitutions as follows. By completing the square in the quadratic trinomial and subsequently substituting $u=x+\frac{b}{2a}$, the original integral is reduced to an integral of one of the following three types:

1) $\int R(u,\sqrt{l^2-u^2})\,du,$

2) $\int R(u, \sqrt{l^2+u^2})\,du,$

3) $\int R(u, \sqrt {u^2-l^2})\,du.$

The latter integrals are transformed by trigonometric or hyperbolic substitution respectively:

1) $u=l\sin t$ или $u=l$th$t,$

2) $u=l$tg$t$ или $u=l$sh$t,$

3) $u=l$sec$t$ или $u=l$ch$t$

to integrals of the form $\int R(\sin t, \cos t),dt$ or $R(\sinh t, \cosh t), dt.$

Integrals of the form $$\int\frac{dx}{(mx+n)^r\sqrt{ax^2+bx+c}} (r=1, 2)$$ are reduced to the integrals discussed above using the substitution $mx+n=\frac{1}{t}$.

Examples.

1. $\int\frac{dx}{(5+x)\sqrt{1+x}}.$

Solution.

We make the substitution $x+1=t^2.$ This yields $$\int\frac{dx}{(5+x)\sqrt{1+x}}=\left[\begin{array}{lcl}x+1=t^2\\t=\sqrt{x+1}\\dt=\frac{1}{2\sqrt{x+1}}dx\end{array}\right]=\int\frac{2dt}{5+t^2-1}=$$ $$=2\int\frac{dt}{t^2+4}=arctg\frac{t}{2}+C=arctg\frac{\sqrt{x+1}}{2}+C.$$

Answer: $arctg\frac{\sqrt{x+1}}{2}+C.$

2.$\int\frac{dx}{\sqrt x-\sqrt[3]{x}}.$

Solution.

Делаем подстановку $x=t^6.$ Получаем $$\int\frac{dx}{\sqrt{x}-\sqrt[3]{x}}=\left[\begin{array}{lcl}x=t^6\\t=\sqrt[6]{x}\\dt=\frac{1}{6\sqrt[6]{x^5}}dx\end{array}\right]=\int\frac{dx}{x^{5/6}(x^{-2/6}-x^{-3/6})}=$$ $$=\int\frac{6dt}{t^{-2}-t^{-3}}=6\int\frac{t^3dt}{t-1}=6\int(t^2+t+1+\frac{1}{t-1})=$$ $$=6\left(\frac{t^3}{3}+\frac{t^2}{2}+t+\ln|t-1|\right)+C=$$ $$=2\sqrt {x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln|\sqrt[6]{x}-1|+C.$$

Answer: $2\sqrt {x}+3\sqrt[3]{x}+6\sqrt[6]{x}+6\ln|\sqrt[6]{x}-1|+C.$

3.$\int\frac{dx}{(\sqrt[3]{x}+4)\sqrt{x}}.$

Solution.

$$\int\frac{dx}{(\sqrt[3]{x}+4)\sqrt{x}}=\int\frac{dx}{x^{5/6}+4x^{1/2}}=\left[\begin{array}{lcl}x=t^6\\t=\sqrt[6]{x}\\dt=\frac{1}{6\sqrt[6]{x^5}}dx\end{array}\right]=$$ $$=\int\frac{dx}{x^{5/6}(1+4x^{-1/3})}=\int\frac{6dt}{1+4t^{-2}}=6\int\frac{t^2dt}{t^2+4}=6\int\frac{t^2+4-4}{t^2+4}dt=$$ $$=6t-12arctg\frac{t}{2}+C=6\sqrt[6]x-12arctg\frac{\sqrt[6]{x}}{2}.$$

Answer: $6\sqrt[6]x-12arctg\frac{\sqrt[6]{x}}{2}.$

4. $\int\frac{dx}{(x^2-3)\sqrt{4-x^2}}.$

Solution.

This is an integral of the second type. To solve it, we'll transform it using substitution into an integral involving trigonometric functions, and then into an integral of a rational function.

$$\int\frac{dx}{(x^2-3)\sqrt{4-x^2}}=\left[\begin{array}{lcl}x=2\sin t\\dx=2\cos t\,dt\end{array}\right]=$$ $$=\int\frac{2\cos tdt}{(4\sin^2 t-3)\sqrt{4-4\sin^2t }}=\int\frac{2\cos t\,dt}{(1-4\cos^2t)2\cos t}=$$ $$\int\frac{dt}{1-4\cos^2 t}=\int\frac{dt}{1-4\frac{1+\cos 2t}{2}}=\int\frac{dt}{-1-2\cos 2t}=$$

$$=\left[\begin{array}{lcl}2t=z\\dz=2\,dt\end{array}\right]=\frac{1}{2}\int\frac{dz}{1+2\cos z}=\left[\begin{array}{lcl}dz=\frac{2dp}{1+p^2}\\\cos z=\frac{1-p^2}{1+p^2}\end{array}\right]=$$ $$=\frac{1}{2}\int\frac{2dp}{(1+p^2)(1+2\frac{1-p^2}{1+p^2})}=\int\frac{dp}{1+p^2+2(1-p^2)}=\int\frac{dp}{3-p^2}=$$ $$=\int\frac{dp}{(\sqrt 3-p)(\sqrt 3+p)}.$$

Let's decompose the integrand into partial fractions:

$$\frac{1}{3-p^2}=\frac{A}{\sqrt 3-p}+\frac{B}{\sqrt 3+p}=\frac{A(\sqrt 3+p)+B(\sqrt 3-p)}{3-p^2}$$ $$=\frac{p(A-B)+\sqrt 3(A+B)}{(3-p^2}$$ Отсюда, приравнивая коэффициенты при одинаковых степенях $p$ находим $A$ и $B.$ $$\left\{\begin{array}{lcl}A-B=0\\\sqrt 3(A+B)=1\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}A=B\\2A=\frac{1}{\sqrt 3}\end{array}\right.\Rightarrow\left\{\begin{array}{lcl}A=\frac{1}{2\sqrt 3}\\B=\frac{1}{2\sqrt 3}\end{array}\right.$$

Thus, $$\int\frac{dp}{3-p^2}=\int\left(\frac{1}{2\sqrt 3}\frac{1}{\sqrt 3+p}+\frac{1}{2\sqrt 3}\frac{1}{\sqrt 3-p}\right)dz=$$ $$=\frac{1}{2\sqrt 3}\left(\ln\left|\sqrt 3+p\right|-\ln\left|\sqrt 3-p\right|\right)+C=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3+p}{\sqrt 3-p}\right|+C.$$ From here, performing the reverse substitutions, we obtain:

$$\int\frac{dx}{(x^2-3)\sqrt{4-x^2}}=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3+p}{\sqrt 3-p}\right|+C=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3+ tg\frac{z}{2}}{\sqrt 3-tg\frac{z}{2}}\right|+C=$$ $$=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3-tg t}{\sqrt 3+tg t}\right|+C=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}}{\sqrt 3\sqrt{1-\frac{x^2}{4}}+\frac{x}{2}}\right|+C=$$ $$=\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3\sqrt{4-x^2}-x}{\sqrt 3\sqrt{4-x^2}+x}\right|+C.$$

Answer: $\frac{1}{2\sqrt 3}\ln\left|\frac{\sqrt 3\sqrt{4-x^2}-x}{\sqrt 3\sqrt{4-x^2}+x}\right|+C.$

5. $\int\frac{\sqrt{x^2-1}}{x}\,dx.$

Solution.

This is an integral of the second type. To solve it, we'll transform it using substitution into an integral involving hyperbolic functions, and then into an integral of a rational function.

$$\int\frac{\sqrt{x^2-1}dx}{x}=\left[\begin{array}{lcl}x=ch t\\dx=sh t\,dt\end{array}\right]=\int\frac{sh^2 tdt}{ch t}=\int\frac{ch^2t-1}{ch t}dt=$$ $$=\int\left(ch t-\frac{1}{ch t}\right)dt=sh t-\int\frac{1}{ch t}dt+C.$$

Let's compute the remaining integral:

$$\int\frac{1}{ch t}dt=\left[\begin{array}{lcl}z=th\frac{t}{2}\\ ch t=\frac{1+z^2}{1-z^2}\\dt=\frac{2dz}{1-z^2}\end{array}\right]=\int\frac{2dz}{(1-z^2)\frac{1+z^2}{1-z^2}}=$$ $$=2\int\frac{dz}{1+z^2}=2 arctg z+C=2arctg th\frac{t}{2}=2arctg\sqrt{\frac{ch t-1}{ch t+1}}+C.$$

Thus,

$$\int\frac{\sqrt{x^2-1}dx}{x}=\sqrt{ch^2 t-1}-2 arctg\sqrt{\frac{ch t-1}{ch t+1}}+C=$$ $$=\sqrt{x^2-1}-2 arctg\sqrt{\frac{x-1}{x+1}}+C.$$

Answer: $\sqrt{x^2-1}-2 arctg\sqrt{\frac{x-1}{x+1}}+C.$

6. $\int\frac{dx}{\sqrt{(x^2+9)}^3}.$

Solution.

This is an integral of the second type. To solve it, we'll transform it using substitution into an integral involving trigonometric functions, and then into an integral of a rational function.

$$\int\frac{dx}{\sqrt{(x^2+9)^3}}=\left[\begin{array}{lcl}x=3 tg t\\dx=\frac{3}{\cos^2 t}\,dt\end{array}\right]=\int\frac{3dt}{\cos^2 t\sqrt{(9 tg^2 t+9)^3}}=\frac{1}{9}\int\frac{dt}{\cos^2 t}=$$ $$=\frac{1}{9}\int\frac{dt}{\cos^2 t\sqrt{\left(\frac{1}{\cos^2 t}\right)^3}}dt=\frac{1}{9}\int\cos t\,dt=\frac{1}{9}\sin t+C=\frac{1}{9}tg t\cdot\cos t+C=$$ $$=\frac{1}{9}tg t\sqrt{\frac{1}{tg^2 t+1}}+C=\frac{1}{9}\frac{x}{3}\sqrt{\frac{1}{\frac{x^2}{9}+1}}+C=\frac{x}{9\sqrt{x^2+9}}+C.$$

Answer: $\frac{x}{9\sqrt{x^2+9}}+C.$

Homework.

1. $\int\frac{xdx}{\sqrt[3]{2x-3}}.$

2. $\int\sqrt[3]{\frac{x+1}{x-1}}\frac{dx}{(x-1)^3}.$

3. $\int\frac{1}{x}\sqrt{\frac{x-1}{x+1}}dx.$

4. $\int\frac{dx}{\sqrt{8x-x^2}}.$

5. $\int\frac{\sqrt{x^2-1}}{x^2}dx.$

6. $\int\sqrt{x^2-2x+10}\,dx.$

Tags: Integration of irrational functions, calculus, integral, mathematical analysis