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Question Number 123313 by aurpeyz last updated on 24/Nov/20

$$\int \frac{2x+3}{3x^2+4x+7}dx$$

Answered by Dwaipayan Shikari last updated on 24/Nov/20

$$\int \frac{2x+3}{3x^2+4x+7}dx$$$$=\frac{1}{3}\int \frac{6x+4}{3x^2+4x+7}+\frac{1}{9}\int \frac{5}{(x^2+\frac{4x}{3}+\frac{7}{3})}dx$$$$=\frac{1}{3}log(3x^2+4x+7)+\frac{5}{9}\int \frac{1}{{(x+\frac{2}{3})}^2+{\left(\sqrt{\frac{7}{3}-\frac{4}{9}}\right)}^2}$$$$=\frac{1}{3}log(3x^2+4x+7)+\frac{5}{3\sqrt{17}}{tan}^{-1}\frac{3x+2}{\sqrt{17}}+C$$

Commented by aurpeyz last updated on 25/Nov/20

$$thankyou$$

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