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Question Number 94609 by  M±th+et+s last updated on 20/May/20

$$\int \frac{x^2-1}{\sqrt{x+1}+\sqrt{2x+3}}dx$$

Answered by mathmax by abdo last updated on 20/May/20

$$\mathrm{I}=\int \frac{{\mathrm{x}}^2-1}{\sqrt{\mathrm{x}+1}+\sqrt{2\mathrm{x}+3}}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{({\mathrm{x}}^2-1)(\sqrt{2\mathrm{x}+3}-\sqrt{\mathrm{x}+1})}{\mathrm{x}+2}\mathrm{dx}$$$$=\int \frac{{\mathrm{x}}^2-1}{\mathrm{x}+2}\sqrt{2x+3}dx-\int \frac{{\mathrm{x}}^2-1}{\mathrm{x}+2}\sqrt{\mathrm{x}+1}\mathrm{dx}=\mathrm{A}-\mathrm{B}\mathrm{changement}$$$$\sqrt{2\mathrm{x}+3}=\mathrm{t}\mathrm{give}2\mathrm{x}+3={\mathrm{t}}^2\Rightarrow \mathrm{x}=\frac{{\mathrm{t}}^2-3}{2}\Rightarrow$$$$\mathrm{A}=\int \frac{\frac{1}{4}{({\mathrm{t}}^2-2)}^2-1}{\frac{{\mathrm{t}}^2-3}{2}+2}\times \mathrm{t}\left(\mathrm{t}\right)\mathrm{dt}=\frac{1}{2}\int \frac{{({\mathrm{t}}^2-2)}^2-4}{{\mathrm{t}}^2-3+4}{\mathrm{t}}^2\mathrm{dt}$$$$=\frac{1}{2}\int \frac{{\mathrm{t}}^2({\mathrm{t}}^4-{4\mathrm{t}}^2)}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{1}{2}\int \frac{{\mathrm{t}}^6-{4\mathrm{t}}^4}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{1}{2}\int \frac{{\mathrm{t}}^4({\mathrm{t}}^2+1)-{5\mathrm{t}}^4}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\frac{1}{2}\int {\mathrm{t}}^4\mathrm{dt}-\frac{5}{2}\int \frac{{\mathrm{t}}^4-1+1}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{{\mathrm{t}}^5}{10}-\frac{5}{2}\int ({\mathrm{t}}^2-1)\mathrm{dt}-\frac{5}{2}\mathrm{arctant}$$$$=\frac{{\mathrm{t}}^5}{10}-\frac{5}{6}{\mathrm{t}}^3+\frac{5}{2}\mathrm{t}-\frac{5}{2}\arctan \left(\mathrm{t}\right)+{\mathrm{c}}_0$$$$=\frac{{\left(\sqrt{2\mathrm{x}+3}\right)}^5}{10}-\frac{5}{6}{\left(\sqrt{2\mathrm{x}+3}\right)}^3+\frac{5}{2}\left(\sqrt{2\mathrm{x}+3}\right)-\frac{5}{2}\arctan \left(\sqrt{2\mathrm{x}+3}\right)+{\mathrm{c}}_0$$$$\mathrm{changement}\sqrt{\mathrm{x}+1}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{t}}^2-1\Rightarrow$$$$\mathrm{B}=\int \frac{{({\mathrm{t}}^2-1)}^2-1}{{\mathrm{t}}^2-1+2}\times \mathrm{t}\left(2\mathrm{t}\right)\mathrm{dt}=2\int \frac{{\mathrm{t}}^2({\mathrm{t}}^4-{2\mathrm{t}}^2)}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=2\int \frac{{\mathrm{t}}^6-{2\mathrm{t}}^4}{{\mathrm{t}}^2+1}\mathrm{dt}=2\int \frac{{\mathrm{t}}^4({\mathrm{t}}^2+1)-{3\mathrm{t}}^4}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=2\int {\mathrm{t}}^4\mathrm{dt}-6\int \frac{{\mathrm{t}}^4}{{\mathrm{t}}^2+1}\mathrm{dt}=\frac{2}{5}{\mathrm{t}}^5-6\int \frac{{\mathrm{t}}^4-1+1}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$=\frac{2}{5}{\mathrm{t}}^5-6\int ({\mathrm{t}}^2-1)\mathrm{dt}-6\arctan \left(\mathrm{t}\right)$$$$=\frac{2}{5}{\mathrm{t}}^5-{2\mathrm{t}}^3+6\mathrm{t}-6\arctan \left(\mathrm{t}\right)+{\mathrm{c}}_1$$$$=\frac{2}{5}{\left(\sqrt{\mathrm{x}+1}\right)}^5-2{\left(\sqrt{\mathrm{x}+1}\right)}^3+6\sqrt{\mathrm{x}+1}-6\arctan \left(\sqrt{\mathrm{x}+1}\right)+{\mathrm{c}}_1$$$$\mathrm{I}=\mathrm{A}-\mathrm{B}$$$$$$

Commented by  M±th+et+s last updated on 20/May/20

$$welldone.$$$$thankyousir$$

Commented by mathmax by abdo last updated on 20/May/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by MJS last updated on 20/May/20

$$\mathrm{I}\mathrm{found}\mathrm{this};\mathrm{integration}\mathrm{is}\mathrm{easy}\mathrm{but}\mathrm{inserting}$$$$\mathrm{at}\mathrm{the}\mathrm{end}\mathrm{is}\mathrm{tricky}...$$$$\int \frac{x^2-1}{\sqrt{x+1}+\sqrt{2x+3}}dx=$$$$[t=\sqrt{2(x+1)}+\sqrt{2x+3}\to dx=\frac{2\sqrt{(x+1)(2x+3)}}{2\sqrt{x+1}+\sqrt{2(2x+3}}dt]$$$$=\frac{\sqrt{2}}{128}\int \frac{{(t-1)}^3{(t+1)}^3(t^2+1)(t^4-18t^2+1)}{t^6((1+\sqrt{2})t^2-1+\sqrt{2})}dt=$$$$=\mathrm{\Sigma }\mathrm{of}\mathrm{the}\mathrm{following}:$$$$\left(1\right)-12(1-\sqrt{2})\int \frac{dt}{t^2+3-2\sqrt{2}}=12\arctan \left((1+\sqrt{2})t\right)$$$$\left(2\right)\frac{2-\sqrt{2}}{128}\int t^{4}dt=\frac{2-\sqrt{2}}{640}{t}^5$$$$\left(3\right)-\frac{50-27\sqrt{2}}{128}\int t^{2}dt=-\frac{50-27\sqrt{2}}{384}{t}^3$$$$\left(4\right)\frac{166-109\sqrt{2}}{64}\int dt=\frac{166-109\sqrt{2}}{64}t$$$$\left(5\right)-\frac{166+109\sqrt{2}}{64}\int \frac{dt}{t^2}=\frac{166+109\sqrt{2}}{64t}$$$$\left(6\right)\frac{50+27\sqrt{2}}{128}\int \frac{dt}{t^4}=-\frac{50+27\sqrt{2}}{384t^3}$$$$\left(7\right)-\frac{2+\sqrt{2}}{128}\int \frac{dt}{t^6}=\frac{2+\sqrt{2}}{640t^5}$$$$=$$$$12\arctan \left((1+\sqrt{2})(\sqrt{2(x+1)}+\sqrt{2x+3})\right)+$$$$-\frac{2}{5}(x^2-3x+11)\sqrt{x+1}+$$$$+\frac{1}{15}(6x^2-17x+51)\sqrt{2x+3}+C$$

Commented by  M±th+et+s last updated on 20/May/20

$$thankyousir.nicesork$$

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