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Question Number 70253 by oyemi kemewari last updated on 02/Oct/19

Commented by mathmax by abdo last updated on 02/Oct/19

$$letI=\int u^2\sqrt{u^2-2}duchangementu=\sqrt{2}ch\left(x\right)give$$$$I=\int 2{ch}^2\left(x\right)\sqrt{2}sh\left(x\right)\sqrt{2}sh\left(x\right)dx$$$$=4\int {ch}^{2}x{sh}^{2}xdx=4\int {\left(ch\left(x\right)sh\left(x\right)\right)}^{2}dx$$$$=4\int {\left(\frac{1}{2}sh\left(2x\right)\right)}^{2}dx=\int {sh}^2\left(2x\right)dx=\int \frac{ch\left(4x\right)-1}{2}dx$$$$=-\frac{x}{2}+\frac{1}{8}sh\left(4x\right)dx+C$$$$=-\frac{x}{2}+\frac{1}{8}\left\{\frac{e^{4x}-e^{-4x}}{2}\right\}+C$$$$=-\frac{x}{2}+\frac{1}{16}(e^{4x}-e^{-4x})+Cbutx=argch\left(\frac{u}{\sqrt{2}}\right)$$$$=ln(\frac{u}{\sqrt{2}}+\sqrt{\frac{u^2}{2}-1})\Rightarrow$$$$I=-\frac{1}{2}ln(\frac{u}{\sqrt{2}}+\sqrt{\frac{u^2}{2}-1})+\frac{1}{16}\{{(\frac{u}{\sqrt{2}}+\sqrt{\frac{u^2}{2}-1})}^4-{(\frac{u}{\sqrt{2}}+\sqrt{\frac{u^2}{2}-1})}^{-4}\}+C$$$$=$$

Commented by oyemi kemewari last updated on 02/Oct/19

thanks you slr

Commented by mathmax by abdo last updated on 03/Oct/19

$$youarewelcome.$$

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