Question

Integration of dx/√x+a + √x+b =?

Answer 1

Dear student,

$$\begin{gathered} LetI=\int \frac{dx}{\sqrt{x+a}+\sqrt{x+b}} \\ =\int \frac{dx}{\sqrt{x+a}+\sqrt{x+b}}\times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}--[onrationali\sin g] \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{{\left(\sqrt{x+a}\right)}^2-{\left(\sqrt{x+b}\right)}^2}dx--[u\sin g{a}^2-b^2=\left(a+b\right)\left(a-b\right)] \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{x+a-x-b}dx \\ =\int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}dx \\ =\frac{1}{a-b}\int \sqrt{x+a}dx-\frac{1}{a-b}\int \sqrt{x+b}dx \\ =I_1-I_2 \\ Solving{I}_1 \\ {I}_1=\frac{1}{a-b}\int \sqrt{x+a}dx \\ Putx+a=u \\ Ondifferentiating \\ dx=du \\ {I}_1=\frac{1}{a-b}\int \sqrt{u}du \\ =\frac{1}{a-b}\left[\frac{u^{{\displaystyle \frac{1}{2}}+1}}{\frac{1}{2}+1}\right] \\ =\frac{1}{a-b}\left[\frac{u^{{\displaystyle \frac{3}{2}}}}{\frac{3}{2}}\right] \\ =\frac{2}{3\left(a-b\right)}{\left[x+a\right]}^{\frac{3}{2}}--\left(1\right)--[undosubstitution] \\ Solving{I}_2 \\ {I}_2=\frac{1}{a-b}\int \sqrt{x+b}dx \\ Putx+b=v \\ Ondifferentiating \\ dx=dv \\ {I}_2=\frac{1}{a-b}\int \sqrt{v}dv \\ =\frac{1}{a-b}\left[\frac{v^{{\displaystyle \frac{1}{2}+1}}}{\frac{1}{2}+1}\right] \\ =\frac{2}{3\left(a-b\right)}\left[v^{\frac{3}{2}}\right] \\ =\frac{2}{3\left(a-b\right)}{\left[x+b\right]}^{{}^{\frac{3}{2}}}--\left(2\right)--[undosubstitution] \\ Thusfrom\left(1\right)and\left(2\right)weget \\ I=\frac{2}{3\left(a-b\right)}{\left[x+a\right]}^{\frac{3}{2}}-\frac{2}{3\left(a-b\right)}{\left[x+b\right]}^{{}^{\frac{3}{2}}}+C \\ =\frac{2{\left(x+a\right)}^{{\displaystyle \frac{3}{2}}}-2{\left(x+b\right)}^{{\displaystyle \frac{3}{2}}}}{3\left(a-b\right)}+C \\ \Rightarrow \boxed{I=\frac{2\left[{\left(x+a\right)}^{\frac{3}{2}}-2{\left(x+b\right)}^{\frac{3}{2}}\right]}{3\left(a-b\right)}+C} \end{gathered}$$
Regards