What is the integral of (3x-1)sqrt(x+6) ?

The integral of (3x-1)sqrt(x+6) is 2x(x+6)^{3/2}-2/3 (x+6)^{3/2}-4/5 (x+6)^{5/2}+C

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integral of (3x-1)sqrt(x+6)

\int (3 x - 1) x + 6 d x

2 x (x + 6)^{\frac{3}{2}} - \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{4}{5} (x + 6)^{\frac{5}{2}} + C

Solution steps

\int (3 x - 1) x + 6 d x

Apply Integration By Parts

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

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

\int 2 (x + 6)^{\frac{3}{2}} d x = \frac{4}{5} (x + 6)^{\frac{5}{2}}

= 2 x (x + 6)^{\frac{3}{2}} - \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{4}{5} (x + 6)^{\frac{5}{2}}

Add a constant to the solution

= 2 x (x + 6)^{\frac{3}{2}} - \frac{2}{3} (x + 6)^{\frac{3}{2}} - \frac{4}{5} (x + 6)^{\frac{5}{2}} + C

ln (\frac{2 x + 1}{x ^{2} - 4})

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

y = e^{x}, y = e^{2 x}, x = ln (2), x = ln (3)

x \to 0 + lim ((1 + 4 x)^{c o t (x)})

x^{2} + 1

What is the integral of (3x-1)sqrt(x+6) ?
The integral of (3x-1)sqrt(x+6) is 2x(x+6)^{3/2}-2/3 (x+6)^{3/2}-4/5 (x+6)^{5/2}+C