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Question Number 152271 by peter frank last updated on 27/Aug/21

$$\int (3\mathrm{x}-2)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$

Answered by qaz last updated on 27/Aug/21

$$\mathrm{A}=\int (3\mathrm{x}-2)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$$$=\frac{3}{2}\int (2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}-\frac{7}{2}\int \sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\mathrm{dx}$$$$={({\mathrm{x}}^2+\mathrm{x}+1)}^{3/2}-\frac{7}{2}\mathrm{B}$$$${(\mathrm{x}+\frac{1}{2})}^2=\frac{3}{4}\tan {}^2\theta$$$$\mathrm{x}=\sqrt{\frac{3}{4}}\tan \theta -\frac{1}{2}$$$$\mathrm{dx}=\sqrt{\frac{3}{4}}\sec {}^2\theta \mathrm{d}\theta$$$$\mathrm{B}=\frac{3}{4}\int {\sec }^3\theta \mathrm{d}\theta$$$$=\frac{3}{4}(\sec \theta \tan \theta -\int {\tan }^2\theta \sec \theta \mathrm{d}\theta )$$$$=\frac{3}{4}(\sec \theta \tan \theta -(\int (\sec {}^2\theta -1)\sec \theta \mathrm{d}\theta )$$$$=\frac{3}{4}(\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\int \sec \theta \mathrm{d}\theta )$$$$=\frac{3}{4}(\frac{1}{2}\sec \theta \tan \theta +\frac{1}{2}\int \frac{\cos \theta \mathrm{d}\theta }{1-\sin {}^2\theta })$$$$=\frac{3}{4}(\frac{1}{2}\sec \theta \tan \theta +\frac{1}{4}\ln \mid \frac{1-\sin \theta }{1+\sin \theta }\mid )+\mathrm{C}$$$$=\frac{1}{4}(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\frac{3}{16}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-\frac{1}{2}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\mathrm{x}+\frac{1}{2}}\mid +\mathrm{C}$$$$\Rightarrow \mathrm{A}={({\mathrm{x}}^2+\mathrm{x}+1)}^{3/2}-\frac{7}{8}(2\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\frac{21}{32}\ln \mid \frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}-\mathrm{x}-\frac{1}{2}}{\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}+\mathrm{x}+\frac{1}{2}}\mid +\mathrm{C}$$

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