Find the integral $\int\frac{x^3}{\left(\sqrt{1+x}\right)^4}dx$

Step-by-step Solution

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Final answer to the problem

$\frac{x^3}{-1-x}+3\ln\left|-1-x\right|-3x+\frac{3}{2}x^2+C_0$
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Step-by-step Solution

How should I solve this problem?

  • Integrate by parts
  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
  • FOIL Method
  • Load more...
Can't find a method? Tell us so we can add it.
1

Rewrite the fraction $\frac{x^3}{\left(\sqrt{1+x}\right)^4}$ inside the integral as the product of two functions: $x^3\frac{1}{\left(\sqrt{1+x}\right)^4}$

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

We can solve the integral $\int x^3\frac{1}{\left(\sqrt{1+x}\right)^4}dx$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
3

First, identify or choose $u$ and calculate it's derivative, $du$

$\begin{matrix}\displaystyle{u=x^3}\\ \displaystyle{du=3x^{2}dx}\end{matrix}$
4

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\frac{1}{\left(\sqrt{1+x}\right)^4}dx}\\ \displaystyle{\int dv=\int \frac{1}{\left(\sqrt{1+x}\right)^4}dx}\end{matrix}$
5

Solve the integral to find $v$

$v=\int\frac{1}{\left(\sqrt{1+x}\right)^4}dx$
6

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

$u=1+x$
7

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

$du=dx$
8

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

$\int\frac{1}{u^{2}}du$
9

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

$\int u^{-2}du$
10

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $-2$

$\frac{u^{-1}}{-1}$
11

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

$\frac{\frac{1}{u}}{-1}$
12

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

$\frac{1}{-\left(1+x\right)}$
13

Now replace the values of $u$, $du$ and $v$ in the last formula

$\frac{x^3}{-1-x}-3\int\frac{x^{2}}{-1-x}dx$
14

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

$\frac{3}{2}x^2-3x+3\ln\left(-1-x\right)$
15

Gather the results of all integrals

$\frac{x^3}{-1-x}+3\ln\left|-1-x\right|-3x+\frac{3}{2}x^2$
16

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

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

Final answer to the problem

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

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Function Plot

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

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5
6
7
8
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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