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1

Solved example of binomial theorem

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

Solve the integral $\int\frac{x^2+3x+1}{\left(x+1\right)^3}dx$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=x+1 \\ du=dx\end{matrix}$
3

Rewriting $x$ in terms of $u$

$x=u-1$
4

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

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

Multiplying polynomials $3$ and $u-1$

$\int\frac{-2+\left(u-1\right)^2+3u}{u^3}du$
6

Split the fraction $\frac{-2+\left(u-1\right)^2+3u}{u^3}$ inside the integral, in two terms with common denominator $u^3$

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

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

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

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int-2u^{-3}du$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$-2\int u^{-3}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$-2\frac{u^{-2}}{-2}$

Simplify the fraction

$u^{-2}$

Substitute $u$ back for it's value, $x+1$

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{1}{\left(x+1\right)^{2}}$
8

The integral $\int\frac{-2}{u^3}du$ results in: $\frac{1}{\left(x+1\right)^{2}}$

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

Split the fraction $\frac{\left(u-1\right)^2+3u}{u^3}$ inside the integral, in two terms with common denominator $u^3$

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

Simplifying

$\int\frac{\left(u-1\right)^2}{u^3}du$

Expand $\left(u-1\right)^2$

$\int\frac{u^2-2u+1}{u^3}du$

Split the fraction $\frac{u^2-2u+1}{u^3}$ inside the integral, in two terms with common denominator $u^3$

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

Simplifying

$\int\frac{1}{u}du$

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

$\ln\left|u\right|$

Substitute $u$ back for it's value, $x+1$

$\ln\left|x+1\right|$
9

The integral $\int\frac{\left(u-1\right)^2+3u}{u^3}du$ results in: $\ln\left|x+1\right|$

$\ln\left|x+1\right|$

Split the fraction $\frac{-2u+1}{u^3}$ inside the integral, in two terms with common denominator $u^3$

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

Simplifying

$\int\frac{-2}{u^{2}}du$

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int-2u^{-2}du$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$-2\int u^{-2}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$2u^{-1}$

Substitute $u$ back for it's value, $x+1$

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{2}{x+1}$
10

The integral $\int\frac{-2u+1}{u^3}du$ results in: $\frac{2}{x+1}$

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

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int u^{-3}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{u^{-2}}{-2}$

Substitute $u$ back for it's value, $x+1$

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{1}{-2\left(x+1\right)^{2}}$
11

The integral $\int\frac{1}{u^3}du$ results in: $\frac{1}{-2\left(x+1\right)^{2}}$

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

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$\int3u^{-2}du$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$3\int u^{-2}du$

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$-3u^{-1}$

Substitute $u$ back for it's value, $x+1$

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

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$\frac{-3}{x+1}$
12

The integral $\int\frac{3}{u^{2}}du$ results in: $\frac{-3}{x+1}$

$\frac{-3}{x+1}$
13

After gathering the results of all integrals, the final answer is

$\frac{1}{\left(x+1\right)^{2}}+\ln\left|x+1\right|+\frac{2}{x+1}+\frac{1}{-2\left(x+1\right)^{2}}+\frac{-3}{x+1}$
14

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

$\frac{1}{\left(x+1\right)^{2}}+\ln\left|x+1\right|+\frac{2}{x+1}+\frac{1}{-2\left(x+1\right)^{2}}+\frac{-3}{x+1}+C_0$

Answer

$\frac{1}{\left(x+1\right)^{2}}+\ln\left|x+1\right|+\frac{2}{x+1}+\frac{1}{-2\left(x+1\right)^{2}}+\frac{-3}{x+1}+C_0$

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