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Question Number 152270 by peter frank last updated on 27/Aug/21

$$\int \frac{5\mathrm{x}+3}{\sqrt{{\mathrm{x}}^2+4\mathrm{x}+10}}\mathrm{dx}$$

Answered by Olaf_Thorendsen last updated on 27/Aug/21

$$\mathrm{F}\left(x\right)=\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx$$$$\mathrm{F}(u-2)=\int \frac{5u-7}{\sqrt{u^2+6}}du$$$$\mathrm{F}(u-2)=5\sqrt{u^2+6}-7\arcsin \left(\frac{u}{\sqrt{6}}\right)+\mathrm{C}$$$$\mathrm{F}\left(x\right)=5\sqrt{x^2+4x+10}-7\arcsin \left(\frac{x+2}{\sqrt{6}}\right)+\mathrm{C}$$$$$$

Commented by peter frank last updated on 27/Aug/21

$$\mathrm{sorry}\mathrm{sir}\mathrm{step}\mathrm{two}\mathrm{explanation}$$$$\mathrm{please}$$

Answered by puissant last updated on 27/Aug/21

$$=5\int \frac{x+\frac{3}{5}}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}\int \frac{2x+\frac{6}{5}}{\sqrt{x^2+4x+10}}dx$$$$=\frac{5}{2}\int \frac{2x+4-\frac{14}{5}}{\sqrt{x^2+4x+10}}dx$$$$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}$$$$=5\int \frac{2x+4}{2\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{{(x+2)}^2+{\left(\sqrt{6}\right)}^2}}$$$$=5\sqrt{x^2+4x+10}-7ln\mid x+2+\sqrt{x^2+4x+10}\mid +C$$$$because\int \frac{dx}{\sqrt{x^2+a^2}}=ln\mid x+\sqrt{x^2+a^2}\mid +C$$$$$$$$\therefore \because Q=5\sqrt{x^2+4x+10}-7ln\mid x+2+\sqrt{x^2+4x+10}\mid +C..$$

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