What is the integral of (x+1)/3 ?

The integral of (x+1)/3 is 1/3 ((x^2)/2+x)+C

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integral of (x+1)/3

\int \frac{x + 1}{3} d x

\frac{1}{3} (\frac{x ^{2}}{2} + x) + C

Solution steps

\int \frac{x + 1}{3} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int 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

= \frac{1}{3} (\int x d x + \int 1 d x)

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

\int 1 d x = x

= \frac{1}{3} (\frac{x ^{2}}{2} + x)

Add a constant to the solution

= \frac{1}{3} (\frac{x ^{2}}{2} + x) + C

x y^{2} d x + (1 + x^{2} y) d y = 0

\int e^{(x^{2} - l n (x))} d x

\frac{( 1 )}{( s ^{2} + 1 ) ^{2}}

\int_{- \frac{π}{2}}^{π} x sin (∣ x ∣) d x

\frac{d}{d x} (arctan (x - 2))