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\int\frac{2x^3+7x^2+2x+9}{2x+3}dx

Integral of (2x^3+7x^2+2x+9)/(2x+3)

Answer

$\frac{x^{3}}{3}+\frac{15}{2}\ln\left|2x+3\right|+x\left(x-2\right)+C_0$

Step-by-step explanation

Problem to solve:

$\int\frac{2x^3+7x^2+2x+9}{2x+3}dx$
1

Divide $2x^3+7x^2+2x+9$ by $2x+3$

$\left|\begin{matrix}2x^{3} & 7x^{2} & 2x & 9 \\ -2x^{3} & -3x^{2} & & \\ & 4x^{2} & 2x & 9 \\ & -4x^{2} & -6x & \\ & & -4x & 9 \\ & & 4x & 6 \\ & & & 15\end{matrix}\right|\begin{matrix}2x+3 \\ x^{2}+2x-2 \\ \\ \\ \\ \\ \\ \end{matrix}$
2

Resulting polynomial

$\int\left(x^{2}+2x-2+\frac{15}{2x+3}\right)dx$

Unlock this step-by-step solution!

Answer

$\frac{x^{3}}{3}+\frac{15}{2}\ln\left|2x+3\right|+x\left(x-2\right)+C_0$
$\int\frac{2x^3+7x^2+2x+9}{2x+3}dx$

Main topic:

Integration by substitution

Used formulas:

8. See formulas

Time to solve it:

~ 1.37 seconds