Integral of (3x-1)/(x+1+2x^2)

\int\frac{3x-1}{2x^2+x+1}dx

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Answer

$-\frac{\sqrt{2}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|-\frac{1}{2}x+x^2+\frac{1}{2}\right|+C_0$

Step by step solution

Problem

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

Split the fraction $\frac{3x+-1}{1+x+2x^2}$ in two terms with same denominator

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

The integral of a sum of two or more functions is equal to the sum of their integrals

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

Taking the constant out of the integral

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

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

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

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

$\int\frac{-1}{2\left(-\frac{1}{16}+\frac{1}{16}+\frac{1}{2}+\frac{1}{2}x+x^2\right)}dx+3\int\frac{x}{1+x+2x^2}dx$
6

Factor the perfect square trinomial $x^2+\frac{1}{2}x+\frac{1}{16}$

$\int\frac{-1}{2\left(-\frac{1}{16}+\frac{1}{2}+\left(x-\frac{1}{4}\right)^2\right)}dx+3\int\frac{x}{1+x+2x^2}dx$
7

Subtract the values $\frac{1}{2}$ and $-\frac{1}{16}$

$\int\frac{-1}{2\left(\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right)}dx+3\int\frac{x}{1+x+2x^2}dx$
8

Taking the constant out of the integral

$\frac{1}{2}\int\frac{-1}{\left(x-\frac{1}{4}\right)^2+\frac{7}{16}}dx+3\int\frac{x}{1+x+2x^2}dx$
9

Solve the integral $\int\frac{-1}{\left(x-\frac{1}{4}\right)^2+\frac{7}{16}}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x-\frac{1}{4} \\ du=dx\end{matrix}$
10

Substituting $u$ and $dx$ in the integral

$\frac{1}{2}\int\frac{-1}{u^2+\frac{7}{16}}du+3\int\frac{x}{1+x+2x^2}dx$
11

Taking the constant out of the integral

$\frac{1}{2}\left(-1\right)\int\frac{1}{u^2+\frac{7}{16}}du+3\int\frac{x}{1+x+2x^2}dx$
12

Multiply $-1$ times $\frac{1}{2}$

$3\int\frac{x}{1+x+2x^2}dx-\frac{1}{2}\int\frac{1}{u^2+\frac{7}{16}}du$
13

Factor the integral's denominator by $\frac{7}{16}$

$3\int\frac{x}{1+x+2x^2}dx-\frac{1}{2}\cdot \frac{16}{7}\int\frac{1}{\frac{16}{7}u^2+1}du$
14

Multiply $\frac{16}{7}$ times $-\frac{1}{2}$

$3\int\frac{x}{1+x+2x^2}dx-\frac{8}{7}\int\frac{1}{\frac{16}{7}u^2+1}du$
15

Solve the integral applying the substitution $v^2=\frac{16}{7}u^2$

$3\int\frac{x}{1+x+2x^2}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}\int\frac{1}{v^2+1}dv$
16

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

$3\int\frac{x}{1+x+2x^2}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}\cdot 1arctan\left(v\right)$
17

Substitute $v$ back for it's value, $\sqrt{2}u$

$3\int\frac{x}{1+x+2x^2}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}u\right)$
18

Substitute $u$ back for it's value, $x-\frac{1}{4}$

$3\int\frac{x}{1+x+2x^2}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
19

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

$3\int\frac{x}{2\left(\frac{1}{2}+\frac{1}{2}x+x^2\right)}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
20

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

$3\int\frac{x}{2\left(-\frac{1}{16}+\frac{1}{16}+\frac{1}{2}+\frac{1}{2}x+x^2\right)}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
21

Factor the perfect square trinomial $x^2+\frac{1}{2}x+\frac{1}{16}$

$3\int\frac{x}{2\left(-\frac{1}{16}+\frac{1}{2}+\left(x-\frac{1}{4}\right)^2\right)}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
22

Subtract the values $\frac{1}{2}$ and $-\frac{1}{16}$

$3\int\frac{x}{2\left(\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right)}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
23

Taking the constant out of the integral

$3\cdot \frac{1}{2}\int\frac{x}{\left(x-\frac{1}{4}\right)^2+\frac{7}{16}}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
24

Multiply $\frac{1}{2}$ times $3$

$\frac{3}{2}\int\frac{x}{\left(x-\frac{1}{4}\right)^2+\frac{7}{16}}dx-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
25

Solve the integral $\int\frac{x}{\left(x-\frac{1}{4}\right)^2+\frac{7}{16}}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x-\frac{1}{4} \\ du=dx\end{matrix}$
26

Rewriting $x$ in terms of $u$

$x=\frac{1}{4}+u$
27

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

$\frac{3}{2}\int\frac{\frac{1}{4}+u}{u^2+\frac{7}{16}}du-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
28

Split the fraction $\frac{u+\frac{1}{4}}{u^2+\frac{7}{16}}$ in two terms with same denominator

$\frac{3}{2}\int\left(\frac{\frac{1}{4}}{u^2+\frac{7}{16}}+\frac{u}{u^2+\frac{7}{16}}\right)du-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
29

The integral of a sum of two or more functions is equal to the sum of their integrals

$\frac{3}{2}\left(\int\frac{\frac{1}{4}}{u^2+\frac{7}{16}}du+\int\frac{u}{u^2+\frac{7}{16}}du\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
30

Taking the constant out of the integral

$\frac{3}{2}\left(\frac{1}{4}\int\frac{1}{u^2+\frac{7}{16}}du+\int\frac{u}{u^2+\frac{7}{16}}du\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
31

Factor the integral's denominator by $\frac{7}{16}$

$\frac{3}{2}\left(\frac{1}{4}\cdot \frac{16}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\int\frac{u}{u^2+\frac{7}{16}}du\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
32

Multiply $\frac{16}{7}$ times $\frac{1}{4}$

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\int\frac{u}{u^2+\frac{7}{16}}du\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
33

Solve the integral $\int\frac{u}{u^2+\frac{7}{16}}du$ applying u-substitution. Let $v$ and $dv$ be

$\begin{matrix}v=u^2+\frac{7}{16} \\ dv=2udu\end{matrix}$
34

Isolate $du$ in the previous equation

$\frac{dv}{2u}=du$
35

Substituting $v$ and $du$ in the integral

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\int\frac{1}{2v}dv\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
36

Taking the constant out of the integral

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\frac{1}{2}\int\frac{1}{v}dv\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
37

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\frac{1}{2}\cdot 1\ln\left|v\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
38

Substitute $v$ back for it's value, $u^2+\frac{7}{16}$

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\frac{1}{2}\ln\left|u^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
39

Substitute $u$ back for it's value, $x-\frac{1}{4}$

$\frac{3}{2}\left(\frac{4}{7}\int\frac{1}{\frac{16}{7}u^2+1}du+\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
40

Solve the integral applying the substitution $v^2=\frac{16}{7}u^2$

$\frac{3}{2}\left(\frac{4}{7}\cdot\frac{\sqrt[3]{3}}{2}\int\frac{1}{v^2+1}dv+\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
41

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

$\frac{3}{2}\left(\frac{4}{7}\cdot\frac{\sqrt[3]{3}}{2}\cdot 1arctan\left(v\right)+\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
42

Substitute $v$ back for it's value, $\sqrt{2}u$

$\frac{3}{2}\left(\frac{4}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}u\right)+\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
43

Substitute $u$ back for it's value, $x-\frac{1}{4}$

$\frac{3}{2}\left(\frac{4}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|\right)-\frac{8}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)$
44

Multiply $\left(\frac{1}{2}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|+\frac{4}{7}\cdot\frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)\right)$ by $\frac{3}{2}$

$-\frac{8}{7}\cdot \frac{\sqrt[3]{3}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|$
45

Multiply $\frac{\sqrt[3]{3}}{2}$ times $-\frac{8}{7}$

$-\frac{\sqrt{2}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|\left(x-\frac{1}{4}\right)^2+\frac{7}{16}\right|$
46

Expanding the polynomial

$-\frac{\sqrt{2}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|-\frac{1}{2}x+x^2+\frac{1}{2}\right|$
47

Add the constant of integration

$-\frac{\sqrt{2}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|-\frac{1}{2}x+x^2+\frac{1}{2}\right|+C_0$

Answer

$-\frac{\sqrt{2}}{2}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{\sqrt[3]{8}}{3}arctan\left(\sqrt{2}\left(x-\frac{1}{4}\right)\right)+\frac{3}{4}\ln\left|-\frac{1}{2}x+x^2+\frac{1}{2}\right|+C_0$