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Find the integral $\int\frac{2x^3}{2x^2+4x+3}dx$

Step-by-step Solution

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Solution

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Final Answer

$\frac{1}{2}x^2-2x+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+C_0$
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Step-by-step Solution

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We could not solve this problem by using the method: Integrate by partial fractions

1

Take out the constant $2$ from the integral

$2\int\frac{x^3}{2x^2+4x+3}dx$
2

Divide $x^3$ by $2x^2+4x+3$

$\begin{array}{l}\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3;}{\phantom{;}\frac{1}{2}x\phantom{;}-1\phantom{;}\phantom{;}}\\\phantom{;}2x^{2}+4x\phantom{;}+3\overline{\smash{)}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3;}\underline{-x^{3}-2x^{2}-\frac{3}{2}x\phantom{;}\phantom{-;x^n}}\\\phantom{-x^{3}-2x^{2}-\frac{3}{2}x\phantom{;};}-2x^{2}-\frac{3}{2}x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}2x^{2}+4x\phantom{;}+3-;x^n;}\underline{\phantom{;}2x^{2}+4x\phantom{;}+3\phantom{;}\phantom{;}}\\\phantom{;\phantom{;}2x^{2}+4x\phantom{;}+3\phantom{;}\phantom{;}-;x^n;}\phantom{;}\frac{5}{2}x\phantom{;}+3\phantom{;}\phantom{;}\\\end{array}$
3

Resulting polynomial

$2\int\left(\frac{1}{2}x-1+\frac{\frac{5}{2}x+3}{2x^2+4x+3}\right)dx$
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

$2\int\frac{1}{2}xdx+2\int-1dx+2\int\frac{\frac{5}{2}x+3}{2x^2+4x+3}dx$
5

The integral $2\int\frac{1}{2}xdx$ results in: $\frac{1}{2}x^2$

$\frac{1}{2}x^2$
6

The integral $2\int-1dx$ results in: $-2x$

$-2x$
7

The integral $2\int\frac{\frac{5}{2}x+3}{2x^2+4x+3}dx$ results in: $\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)$
8

Gather the results of all integrals

$\frac{1}{2}x^2-2x+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)$
9

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$\frac{1}{2}x^2-2x+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+C_0$

Final Answer

$\frac{1}{2}x^2-2x+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+C_0$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of (2x^3/(2x^2+4x))dx using basic integralsSolve integral of (2x^3/(2x^2+4x))dx using u-substitutionSolve integral of (2x^3/(2x^2+4x))dx using integration by partsSolve integral of (2x^3/(2x^2+4x))dx using trigonometric substitution

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Function Plot

Plotting: $\frac{1}{2}x^2-2x+\frac{5}{4}\ln\left(\frac{1}{2}+\left(x+1\right)^2\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+C_0$

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◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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Main Topic: Integrals of Rational Functions

Integrals of rational functions of the form R(x) = P(x)/Q(x).

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