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$\int\frac{x^3-1}{x^2}dx$ is equal to |
$\frac{x^2}{2}+x+c$, where c is constant of integration $\frac{x^2}{2}-\frac{1}{x}+c$, where c is constant of integration $\frac{x^2}{2}-x+c$, where c is constant of integration $\frac{x^2}{2}+\frac{1}{x}+c$, where c is constant of integration |
$\frac{x^2}{2}+\frac{1}{x}+c$, where c is constant of integration |
The correct answer is Option (4) → $\frac{x^2}{2}+\frac{1}{x}+c$, where c is constant of integration Given: $\displaystyle \int \frac{x^{3}-1}{x^{2}}\,dx$ Simplify the integrand: $\frac{x^{3}-1}{x^{2}}=x-\frac{1}{x^{2}}$ Integrate termwise: $\int x\,dx-\int x^{-2}\,dx$ $=\frac{x^{2}}{2}+\frac{1}{x}+C$ Final answer: $\frac{x^{2}}{2}+\frac{1}{x}+C$ |
