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Question Number 128900 by bemath last updated on 11/Jan/21

$$\int \frac{4\mathrm{x}+5}{(\mathrm{x}+2)(\mathrm{x}+3)(\mathrm{x}+4)(\mathrm{x}+5)+1}\mathrm{dx}?$$

Answered by liberty last updated on 11/Jan/21

$$\int \frac{4\mathrm{x}+5}{({\mathrm{x}}^2+7\mathrm{x}+10)({\mathrm{x}}^2+7\mathrm{x}+12)+1}\mathrm{dx}=(\ast )$$$$({\mathrm{x}}^2+7\mathrm{x}+10)({\mathrm{x}}^2+7\mathrm{x}+12)+1$$$$\mathrm{let}{\mathrm{x}}^2+7\mathrm{x}=\mathrm{u}\Rightarrow (\mathrm{u}+10)(\mathrm{u}+12)+1={\mathrm{u}}^2+22\mathrm{u}+121$$$$={(\mathrm{u}+11)}^2={({\mathrm{x}}^2+7\mathrm{x}+11)}^2$$$$(\ast )\int \frac{4\mathrm{x}+5}{{({\mathrm{x}}^2+7\mathrm{x}+11)}^2}\mathrm{dx}=$$$$\int \frac{2(2\mathrm{x}+7)-9}{{({\mathrm{x}}^2+7\mathrm{x}+11)}^2}\mathrm{dx}=\int \frac{2(2\mathrm{x}+7)}{{({\mathrm{x}}^2+7\mathrm{x}+11)}^2}\mathrm{dx}-\int \frac{9}{{({\mathrm{x}}^2+7\mathrm{x}+11)}^2}\mathrm{dx}$$$$=-\frac{2}{{\mathrm{x}}^2+7\mathrm{x}+11}-\int \frac{9}{{({\mathrm{x}}^2+7\mathrm{x}+11)}^2}\mathrm{dx}$$$$\mathrm{next}..$$

Commented by bramlexs22 last updated on 11/Jan/21

$${\mathrm{I}}_2=\int \frac{9\mathrm{dx}}{{({(\mathrm{x}+\frac{7}{2})}^2-{\left(\frac{\sqrt{5}}{2}\right)}^2)}^2}$$$$\mathrm{let}\mathrm{x}+\frac{7}{2}=\frac{\sqrt{5}}{2}\sec \mathrm{t}$$$${\mathrm{I}}_2=\int \frac{\frac{9\sqrt{5}}{2}\sec \mathrm{t}\tan \mathrm{t}}{\frac{25}{4}\tan {}^4\mathrm{t}}\mathrm{dt}$$$${\mathrm{I}}_2=\frac{18\sqrt{5}}{25}\int \frac{\cos {}^2\mathrm{t}}{\sin {}^3\mathrm{t}}\mathrm{dt}$$$${\mathrm{I}}_2=\frac{18\sqrt{5}}{25}(\int {\csc }^3\mathrm{tdt}-\int \csc \mathrm{t}\mathrm{dt})$$$$$$

Answered by Olaf last updated on 11/Jan/21

$$\mathrm{F}\left(x\right)=\int \frac{4x+5}{(x+2)(x+3)(x+4)(x+5)+1}dx$$$$\mathrm{Let}u=x+\frac{7}{2}$$$$\mathrm{F}\left(x\right)=\int \frac{4u-9}{(u-\frac{3}{2})(u-\frac{1}{2})(u+\frac{1}{2})(u+\frac{3}{2})+1}du$$$$\mathrm{F}\left(x\right)=\int \frac{4u-9}{(u^2-\frac{1}{4})(u^2-\frac{9}{4})+1}du$$$$\mathrm{F}\left(x\right)=\int \frac{4u-9}{u^4-\frac{5}{2}{u}^2+\frac{25}{16}}du$$$$\mathrm{F}\left(x\right)=\int \frac{4u-9}{{(u^2-\frac{5}{4})}^2}du$$$$\mathrm{F}\left(x\right)=\int [\frac{4}{u-\frac{5}{4}}-\frac{4}{{(u^2-\frac{5}{4})}^2}]du$$$$\mathrm{F}\left(x\right)=4\ln \mid u-\frac{5}{4}\mid +\frac{8u}{5(u^2-\frac{5}{4})}-\frac{16}{\sqrt{5}}\mathrm{atanh}\left(\frac{2u}{\sqrt{5}}\right)+\mathrm{C}$$$$\mathrm{F}\left(x\right)=4\ln \mid x+\frac{9}{4}\mid +\frac{4(x+7)}{5(x^2+7x+11)}$$$$-\frac{16}{\sqrt{5}}\mathrm{atanh}\left(\frac{2x+7}{\sqrt{5}}\right)+\mathrm{C}$$

Answered by MJS_new last updated on 11/Jan/21

$$\int \frac{4x+5}{(x+2)(x+3)(x+4)(x+5)+1}dx=$$$$=\int \frac{4x+5}{{(x^2+7x+11)}^2}dx=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=\frac{18x+53}{5(x^2+7x+11)}+\frac{18}{5}\int \frac{dx}{x^2+7x+11}=$$$$=\frac{18x+53}{5(x^2+7x+11)}+\frac{36\sqrt{5}}{25}\int (\frac{1}{2x+7-\sqrt{5}}-\frac{1}{2x+7+\sqrt{5}})dx=$$$$=\frac{18x+53}{5(x^2+7x+11)}+\frac{18\sqrt{5}}{25}(\ln \mid 2x+7-\sqrt{5}\mid -\ln \mid 2x+7+\sqrt{5}\mid =$$$$=\frac{18x+53}{5(x^2+7x+11)}+\frac{18\sqrt{5}}{25}\ln \mid \frac{2x+7-\sqrt{5}}{2x+7+\sqrt{5}}\mid +C$$

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