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Question Number 94520 by john santu last updated on 19/May/20

$$\int \sqrt{e^{4x}+1}dx$$

Answered by niroj last updated on 19/May/20

$$\int \sqrt{{\left({\mathrm{e}}^{\mathrm{x}}\right)}^4+1}\mathrm{dx}$$$$\mathrm{Put}{\mathrm{e}}^{\mathrm{x}}=\mathrm{t}$$$${\mathrm{e}}^{\mathrm{x}}.\mathrm{dx}=\mathrm{dt}$$$$\mathrm{dx}=\frac{1}{\mathrm{t}}\mathrm{dt}$$$$=\int \sqrt{{\mathrm{t}}^4+1}.\frac{1}{\mathrm{t}}\mathrm{dt}$$$$=\int \sqrt{{\mathrm{t}}^2({\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2})}\frac{1}{\mathrm{t}}\mathrm{dt}$$$$=\int \mathrm{t}\sqrt{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}\frac{1}{\mathrm{t}}\mathrm{dt}$$$$=\int \sqrt{{(\mathrm{t}+\frac{1}{\mathrm{t}})}^2-2}\mathrm{dt}$$$$=\int \sqrt{{(\mathrm{t}+\frac{1}{\mathrm{t}})}^2-{\left(\sqrt{2}\right)}^2}\mathrm{dt}$$$$=\frac{(\mathrm{t}+\frac{1}{\mathrm{t}})\sqrt{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}}}{2}-\frac{{\left(\sqrt{2}\right)}^2}{2}\log (\mathrm{t}+\frac{1}{\mathrm{t}}+\sqrt{{\mathrm{t}}^2+\frac{1}{{\mathrm{t}}^2}})+\mathrm{C}$$$$=\frac{\frac{{\mathrm{t}}^2+1}{\mathrm{t}}.\frac{\sqrt{{\mathrm{t}}^4+1}}{\mathrm{t}}}{2}-\log (\frac{{\mathrm{t}}^2+1}{\mathrm{t}}+\frac{\sqrt{{\mathrm{t}}^4+1}}{\mathrm{t}})+\mathrm{C}$$$$=\frac{({\mathrm{t}}^2+1)\sqrt{{\mathrm{t}}^4+1}}{{2\mathrm{t}}^2}-\log \left(\frac{{\mathrm{t}}^2+1+\sqrt{{\mathrm{t}}^4+1}}{\mathrm{t}}\right)+\mathrm{C}$$$$=\frac{({\mathrm{e}}^{2\mathrm{x}}+1)\left(\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}\right)}{{2\mathrm{e}}^{2\mathrm{x}}}-\log \left(\frac{{\mathrm{e}}^{2\mathrm{x}}+1+\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}}{{\mathrm{e}}^{\mathrm{x}}}\right)+\mathrm{C}//.$$$$$$$$$$

Commented by i jagooll last updated on 19/May/20

cooll man ������

Commented by peter frank last updated on 19/May/20

$$thankyou$$

Commented by Mr.D.N. last updated on 19/May/20

great��

Commented by niroj last updated on 19/May/20

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Answered by john santu last updated on 19/May/20

$$\mathrm{let}\mathrm{u}={\mathrm{e}}^{4x}\Rightarrow dx=\frac{du}{4u}$$$$\int \frac{\sqrt{u+1}}{4u}du=\frac{1}{4}\int \frac{\sqrt{u+1}}{u}du$$$$setv=\sqrt{u+1}\Rightarrow du=2vdv$$$$\int \frac{1}{4}\frac{2v^2}{v^2-1}dv=\frac{1}{2}\int (1+\frac{1}{v^2-1})dv$$$$=\frac{1}{2}(v+\frac{1}{2}\ln \mid \frac{v-1}{v+1}\mid )+\mathrm{c}$$$$=\frac{1}{2}\sqrt{e^{4x}+1}+\frac{1}{4}\ln \mid \frac{\sqrt{e^{4x}+1}-1}{\sqrt{e^{4x}+1}+1}\mid +\mathrm{c}$$

Commented by i jagooll last updated on 19/May/20

$$\mathrm{waw}....\mathrm{funtastic}$$

Answered by MJS last updated on 19/May/20

$$\mathrm{why}\mathrm{not}\mathrm{all}\mathrm{at}\mathrm{once}?$$$$\int \sqrt{{\mathrm{e}}^{4x}+1}dx=$$$$[t=\sqrt{{\mathrm{e}}^{4x}+1}\to dx=\frac{\sqrt{{\mathrm{e}}^{4x}+1}}{{2\mathrm{e}}^{4x}}dt]$$$$=\frac{1}{2}\int \frac{t^2}{t^2-1}dt=\frac{t}{2}+\frac{1}{4}\ln \frac{t-1}{t+1}=...$$

Answered by mathmax by abdo last updated on 19/May/20

$$\mathrm{I}=\int \sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}=\mathrm{t}\Rightarrow$$$${\mathrm{e}}^{4\mathrm{x}}={\mathrm{t}}^2-1\Rightarrow 4\mathrm{x}=\ln ({\mathrm{t}}^2-1)\Rightarrow \mathrm{x}=\frac{1}{4}\ln ({\mathrm{t}}^2-1)\Rightarrow \mathrm{dx}=\frac{\mathrm{t}}{2({\mathrm{t}}^2-1)}\mathrm{dt}$$$$\Rightarrow \mathrm{I}=\frac{1}{2}\int {\mathrm{t}}^2\times \frac{\mathrm{dt}}{{\mathrm{t}}^2-1}=\frac{1}{2}\int \frac{{\mathrm{t}}^2-1+1}{{\mathrm{t}}^2-1}\mathrm{dt}$$$$=\frac{\mathrm{t}}{2}+\frac{1}{4}\int (\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}=\frac{\mathrm{t}}{2}+\frac{1}{4}\ln \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}\mid +\mathrm{C}$$$$=\frac{1}{2}\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}+\frac{1}{4}\ln \mid \frac{\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}-1}{\sqrt{{\mathrm{e}}^{4\mathrm{x}}+1}+1}\mid +\mathrm{C}$$

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