Final answer to the problem
$\frac{1}{2}x^2-2x+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)+C_0$
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Step-by-step Solution
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Integrate using trigonometric identities Integrate by partial fractions Integrate by substitution Integrate by parts Integrate using tabular integration Integrate by trigonometric substitution Weierstrass Substitution Integrate using basic integrals Product of Binomials with Common Term FOIL Method
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1
Take out the constant $2$ from the integral
$2\int\frac{x^3}{2x^2+4x+3}dx$
2
Divide $x^3$ by $2x^2+4x+3$
$\begin{array}{l}\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3;}{\phantom{;}\frac{1}{2}x\phantom{;}-1\phantom{;}\phantom{;}}\\\phantom{;}2x^{2}+4x\phantom{;}+3\overline{\smash{)}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3;}\underline{-x^{3}-2x^{2}-\frac{3}{2}x\phantom{;}\phantom{-;x^n}}\\\phantom{-x^{3}-2x^{2}-\frac{3}{2}x\phantom{;};}-2x^{2}-\frac{3}{2}x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3-;x^n;}\underline{\phantom{;}2x^{2}+4x\phantom{;}+3\phantom{;}\phantom{;}}\\\phantom{;\phantom{;}2x^{2}+4x\phantom{;}+3\phantom{;}\phantom{;}-;x^n;}\phantom{;}\frac{5}{2}x\phantom{;}+3\phantom{;}\phantom{;}\\\end{array}$
$2\int\left(\frac{1}{2}x-1+\frac{\frac{5}{2}x+3}{2x^2+4x+3}\right)dx$
4
Expand the integral $\int\left(\frac{1}{2}x-1+\frac{\frac{5}{2}x+3}{2x^2+4x+3}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately
$2\int\frac{1}{2}xdx+2\int-1dx+2\int\frac{\frac{5}{2}x+3}{2x^2+4x+3}dx$
Intermediate steps
5
The integral $2\int\frac{1}{2}xdx$ results in: $\frac{1}{2}x^2$
$\frac{1}{2}x^2$
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Intermediate steps
6
The integral $2\int-1dx$ results in: $-2x$
$-2x$
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7
Gather the results of all integrals
$\frac{1}{2}x^2-2x+2\int\frac{\frac{5}{2}x+3}{2x^2+4x+3}dx$
Intermediate steps
8
Rewrite the expression $\frac{\frac{5}{2}x+3}{2x^2+4x+3}$ inside the integral in factored form
$\frac{1}{2}x^2-2x+2\int\frac{\frac{5}{2}x+3}{2\left(\left(x+1\right)^2+\frac{1}{2}\right)}dx$
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Intermediate steps
9
The integral $2\int\frac{\frac{5}{2}x+3}{2\left(\left(x+1\right)^2+\frac{1}{2}\right)}dx$ results in: $-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)$
$-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)$
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10
Gather the results of all integrals
$\frac{1}{2}x^2-2x+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)$
11
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{2}x^2-2x+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)+C_0$
Final answer to the problem
$\frac{1}{2}x^2-2x+\frac{\sqrt{2}}{2}\arctan\left(1.4142012x+1.4142012\right)-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\left(x+1\right)^2+\frac{1}{2}}}\right)+C_0$