How to solve indefinite integral of square root of x divided by x rise to 5 by dx ?

Short answer: indefinite integral of square root of x divided by x rise to 5 by dx is (-2/(7x^3√x))+C. ∫ (√x)/x^5)dx is not particulary hard integral. You can solve it in 11 easy steps. We will walk you through and explain everything. Let's start.

how to solve indefinite integral of square root of x divided by x rise to 5 by dx

Table of contents

Required assumtions

Usually, we have some additional info about function f of (x). In our case:

  1. f(x) belongs to real numbers
  2. f(x) is integrable in that domain
assumtions for indefinite integral of square root of x divided by x rise to 5 by dx

 

Step by step solution of ∫ (√x)/x^5)dx

We will solve ∫ (√x)/x^5)dx in 11 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 u is equal to u rise to 1 divided by n.

step 1

Step 2

In our case u is equal to x and n is equal to 2. Then we may divide numerator by denominator using formula u rise to n divided by u rise to m is equal to u rise to n minus m. In or case u is equal to x, n is equal to a half, m is equal to 5.

step 2

Step 3

Now we have integral of x rise to half minus 5 by dx.

step 3

Step 4

So, in short it is integral of x rise to -4 and a half by dx. After our modifications we may directly use formula, which says integral of u rise to n by du is equal to u rise to n+1 divided by n+1. In our case u is equal to x, n is equal to -4 and a half.

step 4

Step 5

Thus, we have x rise to -4 and a half plus 1 divided by -4 and a half plus one. We must also add constant C belonging to the set of real numbers because our solution is not a single function but a whole class of functions.

step 5

Step 6

So, let’s change -4 and a half plus 1 into improper fractions of the same denominator 2 and we may add their numerators. -9 plus 2 is equal to -7 and our denominator is 2.

step 6

Step 7

Just to be thorough we may put this into our equation.

step 7

Step 8

Then we have x rise to -7 seconds and in denominator we have also -7 seconds. We remember about adding constant C during each step of modification that we do. To make it smoother we can modify further the equation by moving fraction from denominator to the numerator using formula u divided by a divided by b equals to b divided by a and multiplied by u. In our case u is equal to x, a is -7, b is 2. So we have -2 divided by 7 x rise to -7 seconds.

step 8

Step 9

We may also move x rise to -7 seconds to denominator knowing that u rise to -n equals 1 divided by u rise to n.

step 9

Step 10

And now we have -2 divided by 7 multiplied by x rise to 7 seconds. Also x rise to 7 seconds may be expressed smoother as it is x rise to 3 and a half so we can write it as x rise to 3 multiplied by x rise to a half. Then we have x rise to 3 multiplied by square root of x.

step 10

Step 11

We now put it into our equation, and we have -2 in numerator and 7 multiplied by x rise to 3 square roots of x plus C.

step 11

 

What is indefinite integral of square root of x divided by x rise to 5 by dx?

We finally did it: ∫ (√x)/x^5)dx=(-2/(7x^3√x))+C

(√x)/x^5)dx=(-2/(7x^3√x))+C

Full video of how to solve ∫ (√x)/x^5)dx

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.