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Question Number 74346 by mathmax by abdo last updated on 22/Nov/19

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

Answered by MJS last updated on 24/Nov/19

$$\int \frac{x+\sqrt{x+1}}{3+2\sqrt{x-1}}dx=$$$$[t=2\sqrt{x-1}\to dx=\sqrt{x-1}dt]$$$$=\frac{1}{4}\int \frac{t\sqrt{t^2+8}}{t+3}dt+\frac{1}{8}\int t^{2}dt-\frac{3}{8}\int tdt+\frac{13}{8}\int dt-\frac{39}{8}\int \frac{dt}{t+3}=$$$$=\frac{1}{4}\int \frac{t\sqrt{t^2+8}}{t+3}dt+\frac{1}{24}{t}^3-\frac{3}{16}{t}^2+\frac{13}{8}t-\frac{39}{8}\ln (t+3)$$$$$$$$\int \frac{t\sqrt{t^2+8}}{t+3}dt=$$$$[u=\frac{\sqrt{2}}{4}(t+\sqrt{t^2+8})\to dt=\frac{2\sqrt{2(t^2+8)}}{t+\sqrt{t^2+8}}du]$$$$=-\frac{51\sqrt{2}}{8}\int \frac{du}{u^2+\frac{3\sqrt{2}}{2}u-1}+\frac{1}{2}\int udu-\frac{3\sqrt{2}}{4}\int du+\frac{13}{4}\int \frac{du}{u}+\frac{3\sqrt{2}}{4}\int \frac{du}{u^2}+\frac{1}{2}\int \frac{du}{u^3}=$$$$=\frac{3\sqrt{17}}{4}\ln \frac{4u+(3+\sqrt{17})\sqrt{2}}{4u+(3-\sqrt{17})\sqrt{2}}+\frac{1}{4}{u}^2-\frac{3\sqrt{2}}{4}u+\frac{13}{4}\ln u-\frac{3\sqrt{2}}{4u}-\frac{1}{4u^2}=$$$$...$$$$\mathrm{now}\mathrm{just}\mathrm{insert}\mathrm{the}\mathrm{substitutions}$$

Commented by abdomathmax last updated on 24/Nov/19

$$thankxsirmjs.$$

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