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

Step-by-step Solution

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Solution

Final Answer

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

Specify the solving method

1

Simplify $\left(\sqrt{1+x}\right)^4$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $\frac{1}{2}$ and $n$ equals $4$

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

Expand

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

Divide $x^3$ by $1+2x+x^2$

$\begin{array}{l}\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}{\phantom{;}x\phantom{;}-2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+2x\phantom{;}+1\overline{\smash{)}\phantom{;}x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1;}\underline{-x^{3}-2x^{2}-x\phantom{;}\phantom{-;x^n}}\\\phantom{-x^{3}-2x^{2}-x\phantom{;};}-2x^{2}-x\phantom{;}\phantom{-;x^n}\\\phantom{\phantom{;}x^{2}+2x\phantom{;}+1-;x^n;}\underline{\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}}\\\phantom{;\phantom{;}2x^{2}+4x\phantom{;}+2\phantom{;}\phantom{;}-;x^n;}\phantom{;}3x\phantom{;}+2\phantom{;}\phantom{;}\\\end{array}$
4

Resulting polynomial

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

Simplify the expression inside the integral

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

We can solve the integral $\int\frac{3}{1+x}dx$ 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

$\int xdx+\int-2dx+\int\frac{3}{u}du+\int\frac{-1}{\left(1+x\right)^{2}}dx$
9

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

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

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

$-2x$
11

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

$3\ln\left(1+x\right)+\frac{1}{1+x}$
12

Gather the results of all integrals

$\frac{1}{2}x^2-2x+\frac{1}{1+x}+3\ln\left(1+x\right)$
13

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

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

Final Answer

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

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of ((x^3)/((1+x)^0.5^4))dx using partial fractionsSolve integral of ((x^3)/((1+x)^0.5^4))dx using basic integralsSolve integral of ((x^3)/((1+x)^0.5^4))dx using integration by partsSolve integral of ((x^3)/((1+x)^0.5^4))dx using trigonometric substitution

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Function Plot

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

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5
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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
D□x
∫
∫◻
|◻|
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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