Integration of irrational functions

Integrals of the form $$\int R(x,{\left(\frac{ax+b}{cx+d}\right)}^{\frac{m_1}{n_1}},{\left(\frac{ax+b}{cx+d}\right)}^{\frac{m_2}{n_2}},...)dx$$ 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 $$\int R(x,\sqrt{a{x}^2+bx+c})dx,$$ 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 $$\int \frac{dx}{(mx+n{)}^r\sqrt{a{x}^2+bx+c}}(r=1,2)$$ 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 $$\int \frac{dx}{(5+x)\sqrt{1+x}}=\left[\begin{array}{l}x+1=t^2\\ t=\sqrt{x+1}\\ dt=\frac{1}{2\sqrt{x+1}}dx\end{array}\right]=\int \frac{2dt}{5+t^2-1}=$$ $$=2\int \frac{dt}{t^2+4}=arctg\frac{t}{2}+C=arctg\frac{\sqrt{x+1}}{2}+C.$$

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

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

Solution.

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

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.

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

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.

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

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

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 $B.$ $$\{\begin{array}{l}A-B=0\\ \sqrt{3}(A+B)=1\end{array}\Rightarrow \{\begin{array}{l}A=B\\ 2A=\frac{1}{\sqrt{3}}\end{array}\Rightarrow \{\begin{array}{l}A=\frac{1}{2\sqrt{3}}\\ B=\frac{1}{2\sqrt{3}}\end{array}$$

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

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

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.

$$\int \frac{\sqrt{x^2-1}dx}{x}=\left[\begin{array}{l}x=cht\\ dx=shtdt\end{array}\right]=\int \frac{s{h}^{2}tdt}{cht}=\int \frac{c{h}^{2}t-1}{cht}dt=$$ $$=\int (cht-\frac{1}{cht})dt=sht-\int \frac{1}{cht}dt+C.$$

Let's compute the remaining integral:

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

Thus,

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

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

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.

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

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