What is the integral of (2x^{4/3}+3x-1) ?

The integral of (2x^{4/3}+3x-1) is 6/7 x^{7/3}+(3x^2)/2-x+C

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integral of (2x^{4/3}+3x-1)

\int (2 x^{\frac{4}{3}} + 3 x - 1) d x

\frac{6}{7} x^{\frac{7}{3}} + \frac{3 x ^{2}}{2} - x + C

Solution steps

\int 2 x^{\frac{4}{3}} + 3 x - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x^{\frac{4}{3}} d x + \int 3 x d x - \int 1 d x

\int 2 x^{\frac{4}{3}} d x = \frac{6}{7} x^{\frac{7}{3}}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 1 d x = x

= \frac{6}{7} x^{\frac{7}{3}} + \frac{3 x ^{2}}{2} - x

Add a constant to the solution

= \frac{6}{7} x^{\frac{7}{3}} + \frac{3 x ^{2}}{2} - x + C

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What is the integral of (2x^{4/3}+3x-1) ?
The integral of (2x^{4/3}+3x-1) is 6/7 x^{7/3}+(3x^2)/2-x+C