How to solve indefinite integral of 1 divided by 4th root of x rise to three by dx ?
Short answer: indefinite integral of 1 divided by 4th root of x rise to three by dx is 4∜x+C. ∫ 1/(∜x^3)dx is not particulary hard integral. You can solve it in 10 easy steps. We will walk you through and explain everything. Let's start.
Table of contents
- Required assumtions
- Step by step solution of ∫ 1/(∜x^3)dx
- What is indefinite integral of 1 divided by 4th root of x rise to three by dx?
- Full video how to solve ∫ 1/(∜x^3)dx
- Dictionary
- Want to solve next one?
Required assumtions
Usually, we have some additional info about function f of (x). In our case:
- f(x) belongs to real numbers
- f(x) is integrable in that domain
Step by step solution of ∫ 1/(∜x^3)dx
We will solve ∫ 1/(∜x^3)dx in 10 easy steps. Let's get started
Step 1
From symbol dx we know that differential of variable x indicates that the variable of integration is x. Let’s rewrite integrand - function inside the integral without fraction. Knowing that n root of s is equal to s rise to 1 divided by n. In our case s is equal to x rise to 3 and n is equal to 4.
Step 2
And inside our integral we have now 1 divided by x rise to 3 and the whole x rise to 3 is rise once more to one fourth. Let’s modify a little our denominator. We may use formula that enables multiplication of powers in our case so in case we have s rise to n and once more the the whole thing rise to m, we may write it as s rise to n multiplied by m. In our case s is equal to x, n is 3 and m is one fourth.
Step 3
Now we have integral of one divided by x rise to three fourths by dx.
Step 4
We may avoid also having fraction and x rise to three fourths in denominator if we use formula which says the one divided by s rise to n equals s rise to -n.
Step 5
Then we will have x rise to minus three thirds inside our integral. After these modifications we can now use directly one of formulas to solve the integral. Integral of s rise to n by ds equals s rise to n+1 divided by n+1 plus constant C.
Step 6
Thus, we may write x rise to minus three fourths plus 1 in numerator and minus three fourths plus one in denominator of our solution. We also have to remember to add constant C as our solution is not a single function but a whole class of functions.
Step 7
Let’s add minus three fourths to 1. To do so we may swap 1 into four fourths as we want to have the same denominator of 4. Then we just add minus 3 plus 4 in numerator and finally we have one fourth as a result.
Step 8
So, we put one fourth to our solution and we have now x rise to one fourth divided by one fourth plus constant C. Now we may take advantage from the fact that s divided by a divided by b equals b divided by a and multiplied by s. In our case s is equal to x rise to one fourth, a is 1 and b is 4.
Step 9
And our main solution will be 4x rise to one fourth plus constant C.
Step 10
We may also smoot a little our result taking advantage of formula which says s rise to 1 divided by n equals n-th roots of s. And we have four fourth roots of x plus constant C.
What is indefinite integral of 1 divided by 4th root of x rise to three by dx?
We finally did it: ∫ 1/(∜x^3)dx=4∜x+C
Dictionary
Integration (antidifferentiation)
Computation (process of finding) of an integral, opposite process to differentiation.Integrand
Function placed between sign of integral and differential of variable of integration e.g. $${{ \int f(x)dx}}$$,where:
$${{ \int }}$$- integration operator,
f(x) – integrand,
dx- differential of variable of integration x
Integrable function
Function that integral over its domain is finite.Indefinite integral
Represents a class of primitive functions whose derivative is the integrand e.g. $${{\int f(x)dx=F(x)+C \Leftrightarrow F’(x)=f(x)}}$$$${{C=const. }}$$,
$${{f, F,C \in R }}$$
R-real numbers
$${{\int f(x)dx}}$$ - indefinite integral of function f(x) by dx,
$${{F(x)+C}}$$ – a class of primitive functions that $${{F’(x)=f(x) }}$$,
F(x) - primitive function, usually written in capital letters,
R-real numbers.
Function
Function specified on a set X and having values in set Y is an assignment each element of set X specifically one element in set Y.$${{f: X \rightarrow Y}}$$
f-function name,
X-set of elements of function f, domain of a function f
Y-set of function values of function f, codomain of a function f
$${{x\in X, y \in Y}}$$
$${{y=f(x) f: x\rightarrow y}}$$
$${{y=y(x), }}$$
y(x)-vales of the function named y,
x-independent variable,
y-dependent variable.
