Integrate \; 3/x + 4x² - 3x + 8

3 x - \; 4/3 x³ + \; 3/2 x² - 8x + c

3 x + \; 4/3 x³ - \; 3/2 x² - 8x + c

3 x + \; 4/3 x³ + \; 3/2 x² + 8x + c

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Integral Calculus

Question : 52 of 73

Marks: +1, -0

Integrate

\; \frac{3}{x} + 4x^{2} - 3x + 8

Solution:

Concept:

Integrate term by term using standard integration formulas.

Explanation:

\int \left( \frac{3}{x} + 4x^2 - 3x + 8 \right) dx = \int \frac{3}{x} dx + \int 4x^2 dx - \int 3x dx + \int 8 dx

\int \frac{3}{x} dx = 3 \log |x| + c

\int 4x^2 dx = \frac{4}{3} x^3 + c

\int -3x dx = -\frac{3}{2} x^2 + c

\int 8 dx = 8x + c

Combining:

3 \log |x| + \frac{4}{3}x^3 - \frac{3}{2}x^2 + 8x + c

None of the given options match this result exactly (sign errors in each).

Answer:

3 \log |x| + \frac{4}{3}x^3 - \frac{3}{2}x^2 + 8x + c

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