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$\int \frac{x^3-3 x+7}{x^2+4} d x$ is equal to : |
$\frac{x^2}{2}+\frac{7}{2} \ln \left(x^2+4\right)+c$ $\frac{x^2}{2}+\frac{7}{2} \tan ^{-1} \frac{x}{2}-\frac{7}{2} \ln \left(x^2+4\right)+c$ $-\frac{x^2}{2}+\frac{7}{2} \tan ^{-1} \frac{x^2}{2}+\frac{7}{2} \ln \left(x^2+4\right)+c$ $\frac{x}{2}+\frac{7}{2} \tan ^{-1} \frac{x}{2}+c$ |
$\frac{x^2}{2}+\frac{7}{2} \tan ^{-1} \frac{x}{2}-\frac{7}{2} \ln \left(x^2+4\right)+c$ |
$\frac{x^3-3 x+7}{x^2+4}=x-\frac{7(x-1)}{x^2+4}$ $\int \frac{x^3-3 x+7}{x^2+4} d x=\frac{x^2}{2}-7 \int \frac{(x-1)}{x^2+4} d x$ $=\frac{x^2}{2}+\frac{7}{2} \tan ^{-1} \frac{x}{2}-\frac{7}{2} \ln \left(x^2+4\right)+c$ Hence (2) is the correct answer. |
