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

Step-by-step Solution

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Final Answer

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

Specify the solving method

Use the complete the square method to factor the trinomial of the form $ax^2+bx+c$. Take common factor $a$ ($2$) to all terms

$\frac{2x^3}{2\left(x^2+2x+\frac{3}{2}\right)}$

Add and subtract $\displaystyle\left(\frac{b}{2a}\right)^2$

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

Factor the perfect square trinomial $x^2+2xx+1$

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

Subtract the values $\frac{3}{2}$ and $-1$

$\frac{2x^3}{2\left(\frac{1}{2}+\left(x+1\right)^2\right)}$
1

Rewrite the expression $\frac{2x^3}{2x^2+4x+3}$ inside the integral in factored form

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

Simplify the fraction $\frac{2x^3}{2\left(\frac{1}{2}+\left(x+1\right)^2\right)}$ by $2$

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

A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$

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

Add the values $\frac{1}{2}$ and $1$

$\frac{3}{2}+x^2+2x$
3

Expand

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

Divide $x^3$ by $\frac{3}{2}+x^2+2x$

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

Resulting polynomial

$\int\left(x-2+\frac{\frac{5}{2}x+3}{\frac{1}{2}+\left(x+1\right)^2}\right)dx$
6

Expand the integral $\int\left(x-2+\frac{\frac{5}{2}x+3}{\frac{1}{2}+\left(x+1\right)^2}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately

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

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

$\frac{1}{2}x^2$
7

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

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

The integral of a constant is equal to the constant times the integral's variable

$-2x$
8

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

$-2x$

We can solve the integral $\int\frac{\frac{5}{2}x+3}{\frac{1}{2}+\left(x+1\right)^2}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $x+1$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=x+1$

Now, in order to rewrite $dx$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=dx$

Rewriting $x$ in terms of $u$

$x=u-1$

Substituting $u$, $dx$ and $x$ in the integral and simplify

$\int\frac{\frac{1}{2}+\frac{5}{2}u}{\frac{1}{2}+u^2}du$

Expand the fraction $\frac{\frac{1}{2}+\frac{5}{2}u}{\frac{1}{2}+u^2}$ into $2$ simpler fractions with common denominator $\frac{1}{2}+u^2$

$\int\left(\frac{\frac{1}{2}}{\frac{1}{2}+u^2}+\frac{\frac{5}{2}u}{\frac{1}{2}+u^2}\right)du$

Simplify the expression inside the integral

$\int\frac{1}{2\left(\frac{1}{2}+u^2\right)}du+\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$

Take the constant $\frac{1}{2}$ out of the integral

$\frac{1}{2}\int\frac{1}{\frac{1}{2}+u^2}du+\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$

Solve the integral by applying the formula $\displaystyle\int\frac{x'}{x^2+a^2}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)$

$\frac{1}{2}\cdot \left(\frac{1}{\sqrt{\left(\frac{1}{2}\right)}}\right)\arctan\left(\frac{u}{\sqrt{\left(\frac{1}{2}\right)}}\right)+\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$

Simplify the expression inside the integral

$\frac{\sqrt{2}}{2}\arctan\left(1.414201u\right)+\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$

Replace $u$ with the value that we assigned to it in the beginning: $x+1$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$

We can solve the integral $\frac{5}{2}\int\frac{u}{\frac{1}{2}+u^2}du$ by applying integration method of trigonometric substitution using the substitution

$u=\frac{\sqrt{2}}{2}\tan\left(\theta \right)$

Now, in order to rewrite $d\theta$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=\frac{\sqrt{2}}{2}\sec\left(\theta \right)^2d\theta$

Substituting in the original integral, we get

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\frac{\frac{1}{2}\tan\left(\theta \right)\sec\left(\theta \right)^2}{\frac{1}{2}+\frac{1}{2}\tan\left(\theta \right)^2}d\theta$

Simplifying

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{2\left(\frac{1}{2}+\frac{1}{2}\tan\left(\theta \right)^2\right)}d\theta$

Factor the polynomial $\left(\frac{1}{2}+\frac{1}{2}\tan\left(\theta \right)^2\right)$ by it's greatest common factor (GCF): $\frac{1}{2}$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{1+\tan\left(\theta \right)^2}d\theta$

Applying the trigonometric identity: $1+\tan\left(\theta \right)^2 = \sec\left(\theta \right)^2$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{\sec\left(\theta \right)^2}d\theta$
Why is tan(x)^2+1 = sec(x)^2 ?

Simplify the fraction $\frac{\tan\left(\theta \right)\sec\left(\theta \right)^2}{\sec\left(\theta \right)^2}$ by $\sec\left(\theta \right)^2$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)+\frac{5}{2}\int\tan\left(\theta \right)d\theta$

The integral of the tangent function is given by the following formula, $\displaystyle\int\tan(x)dx=-\ln(\cos(x))$

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)-\frac{5}{2}\ln\left(\cos\left(\theta \right)\right)$

Express the variable $\theta$ in terms of the original variable $x$

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

Replace $u$ with the value that we assigned to it in the beginning: $x+1$

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

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

$\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{2}+\left(x+1\right)^2}}\right)$
10

Gather the results of all integrals

$\frac{1}{2}x^2-2x-\frac{5}{2}\ln\left(\frac{\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{2}+\left(x+1\right)^2}}\right)+\frac{\sqrt{2}}{2}\arctan\left(1.414201\left(x+1\right)\right)$
11

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

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

Final Answer

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

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Function Plot

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

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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
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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