What is the integral of (x+1)(6x-5) ?

The integral of (x+1)(6x-5) is 2x^3+3x^2-(5x^2)/2-5x+C

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integral of (x+1)(6x-5)

\int (x + 1) (6 x - 5) d x

2 x^{3} + 3 x^{2} - \frac{5 x ^{2}}{2} - 5 x + C

Solution steps

\int (x + 1) (6 x - 5) d x

Apply Integration By Parts

: (6 x - 5) (\frac{x ^{2}}{2} + x) - \int 6 (\frac{x ^{2}}{2} + x) d x

= (6 x - 5) (\frac{x ^{2}}{2} + x) - \int 6 (\frac{x ^{2}}{2} + x) d x

\int 6 (\frac{x ^{2}}{2} + x) d x = x^{3} + 3 x^{2}

= (6 x - 5) (\frac{x ^{2}}{2} + x) - (x^{3} + 3 x^{2})

Simplify

(6 x - 5) (\frac{x ^{2}}{2} + x) - (x^{3} + 3 x^{2}) : 2 x^{3} + 3 x^{2} - \frac{5 x ^{2}}{2} - 5 x

= 2 x^{3} + 3 x^{2} - \frac{5 x ^{2}}{2} - 5 x

Add a constant to the solution

= 2 x^{3} + 3 x^{2} - \frac{5 x ^{2}}{2} - 5 x + C

\frac{7 x ^{2}}{4 e ^{x} - x}

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

5 x^{- 2}

\frac{d}{d x} (se x^{2} (x))

\int 5 tan (x) sec^{7} (x) d x

What is the integral of (x+1)(6x-5) ?
The integral of (x+1)(6x-5) is 2x^3+3x^2-(5x^2)/2-5x+C