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Question Number 85441 by jagoll last updated on 22/Mar/20

$$\int \frac{{6\mathrm{x}}^4-4}{\sqrt{{\mathrm{x}}^4-2}}\mathrm{dx}=?$$

Answered by john santu last updated on 22/Mar/20

$$I=\int \frac{4x^4}{\sqrt{x^4-2}}dx=4\int x\left(\frac{x^3}{\sqrt{x^4-2}}\right)dx$$$$=4\int \frac{1}{4}x\frac{d(x^4-2)}{\sqrt{x^4-2}}$$$$=4(\frac{1}{4}x\left(2\sqrt{x^4-2}\right)-\int \frac{1}{2}\sqrt{x^4-2})+c$$$$=2x\sqrt{x^4-2}-2\int \sqrt{x^4-2}dx$$$$J=\int \frac{2x^4-4}{\sqrt{x^4-2}}dx=2\int \frac{x^4-2}{\sqrt{x^4-2}}dx$$$$J=2\int \sqrt{x^4-2}dx$$$$theanswer$$$$I+J=2x\sqrt{x^4-2}-2\int \sqrt{x^4-2}dx+$$$$2\int \sqrt{x^4-2}dx=2x\sqrt{x^4-2}+c.\therefore$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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