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Exercise

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

Step-by-step Solution

1

Solve the product $5\left(1+x^2\right)$

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

Rewrite the expression $\frac{1}{-2x+5+5x^2}$ inside the integral in factored form

$\int\frac{1}{5\left(\left(x-\frac{1}{5}\right)^2+\frac{24}{25}\right)}dx$
3

Take the constant $\frac{1}{5}$ out of the integral

$\frac{1}{5}\int\frac{1}{\left(x-\frac{1}{5}\right)^2+\frac{24}{25}}dx$
4

We can solve the integral $\int\frac{1}{\left(x-\frac{1}{5}\right)^2+\frac{24}{25}}dx$ 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 $x-\frac{1}{5}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

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$u=x-\frac{1}{5}$
5

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 finding the derivative of the equation above

$du=dx$
6

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

$\frac{1}{5}\int\frac{1}{u^2+\frac{24}{25}}du$
7

Factor the integral's denominator by $\frac{24}{25}$

$\frac{1}{5}\cdot \frac{1}{\frac{24}{25}}\int\frac{1}{1+\frac{u^2}{\frac{24}{25}}}du$
8

Simplify the expression

$\frac{5}{24}\int\frac{1}{1+\frac{25u^2}{24}}du$
9

Solve the integral applying the substitution $v^2=\frac{25u^2}{24}$. Then, take the square root of both sides, simplifying we have

$v=\frac{5u}{\sqrt{24}}$
10

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

$dv=\frac{5}{\sqrt{24}}du$
11

Isolate $du$ in the previous equation

$\frac{dv}{\frac{5}{\sqrt{24}}}=du$
12

After replacing everything and simplifying, the integral results in

$\frac{\sqrt{24}}{24}\int\frac{1}{1+v^2}dv$
13

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

$\frac{\sqrt{24}}{24}\arctan\left(v\right)$
14

Multiplying the fraction by $\arctan\left(v\right)$

$\frac{\sqrt{24}\arctan\left(v\right)}{24}$
15

Replace $v$ with the value that we assigned to it in the beginning: $\frac{5u}{\sqrt{24}}$

$\frac{\sqrt{24}\arctan\left(\frac{5u}{\sqrt{24}}\right)}{24}$
16

Replace $u$ with the value that we assigned to it in the beginning: $x-\frac{1}{5}$

$\frac{\sqrt{24}\arctan\left(\frac{5\left(x-\frac{1}{5}\right)}{\sqrt{24}}\right)}{24}$
17

Simplify the product by distributing $5$ to both terms

$\frac{\sqrt{24}\arctan\left(\frac{-1+5x}{\sqrt{24}}\right)}{24}$
18

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

$\frac{\sqrt{24}\arctan\left(\frac{-1+5x}{\sqrt{24}}\right)}{24}+C_0$

Final answer to the exercise

$\frac{\sqrt{24}\arctan\left(\frac{-1+5x}{\sqrt{24}}\right)}{24}+C_0$

Try other ways to solve this exercise

  • Choose an option
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • Load more...
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