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Integration of Root x
Integration of root x is determined using the formula of the integration given by ∫x^{n} dx = x^{n+1}/(n + 1) + C. Since root x is a radical, therefore we substitute n = 1/2 in the formula to get the integration of root x. The integral of square root x is equal to twothird of x raised to the power of three by two plus the integration constant. Mathematically, the integration of root x is written as ∫√x dx = (2/3) x^{3/2} + C.
Further in this article, we will derive the integral of square root x using the integration formula, and the integration of root x square plus a square and the integral of root x square minus a square. We will also solve solve some examples for a better understanding.
1.  What is Integration of Root x? 
2.  Integral of Square Root x Proof 
3.  Definite Integration of Root x 
4.  Integration of Root x Square Plus a Square 
5.  FAQs on Integration of Root x 
What is Integration of Root x?
The integration of root x is nothing but the integral of square root x with respect to x which is given by ∫√x dx = (2/3) x^{3/2} + C, where C is the constant of integration, ∫ denote the symbol of the integration and dx indicates the integral of square root x with respect to x. We can compute the integration of root x using the formula of integration ∫x^{n} dx = x^{n+1}/(n + 1) + C. Here we have n = 1/2 as root x is a radical expression. Hence, the formula for the integration of root x is given by, ∫√x dx = (2/3) x^{3/2} + C, where C is the constant of integration.
Integral of Square Root x Proof
Now that we know that the integration of root x is equal to (2/3) x^{3/2} + C, we will now prove it using the formula of integration. We will use the formula ∫x^{n} dx = x^{n+1}/(n + 1) + C by substituting n = 1/2 in it as √x is nothing but x raised to the power of one by two, that is, √x = x^{1/2}. Therefore, we have
∫√x dx = ∫x^{1/2} dx
= x^{1/2 + 1}/(1/2 + 1) + C
= x^{3/2}/(3/2) + C
= (2/3) x^{3/2} + C
Hence, the integral of square root x is equal to (2/3) x^{3/2} + C, where C is the integration constant.
Definite Integration of Root x
Next, we will find the definite integration of root x with limits from 1 to 10. We know that the integral of the square root x formula is ∫√x dx = (2/3) x^{3/2} + C. In this, we will substitute the limits to find its definite integral.
\(\begin{align} \int_{1}^{10}\sqrt{x} \ dx&= \left [ \frac{2}{3}x^{\frac{3}{2}} \right ]_1^{10}\\&=\frac{2}{3}(10)^{\frac{3}{2}}  \frac{2}{3}(1)^{\frac{3}{2}}\\&=\frac{2}{3}[10^{\frac{3}{2}}1]\end{align}\)
Integration of Root x Square Plus a Square
In this section, we will find the integral of the square root of x square plus a square, that is, √(x^{2} + a^{2}). To find this integral, we will use the method of integration by parts and the formula ∫1/√(x^{2} + a^{2}) dx = log x + √(x^{2} + a^{2}) + C. The formula for integration by parts is ∫f(x) g(x) dx = f(x) ∫g(x) dx  ∫[d(f(x))/dx ∫ g(x) dx] dx. Here f(x) = √(x^{2} + a^{2}) and g(x) = 1 as we can write √(x^{2} + a^{2}) as √(x^{2} + a^{2}).1. Hence, we have
∫√(x^{2} + a^{2}) dx = ∫√(x^{2} + a^{2}).1 dx
⇒ ∫√(x^{2} + a^{2}) dx = √(x^{2} + a^{2}) ∫dx  ∫[d(√(x^{2} + a^{2}))/dx ∫dx] dx
⇒ ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2})  ∫x^{2}/√(x^{2} + a^{2}) dx
⇒ ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2})  ∫(a^{2}  a^{2} + x^{2})/√(x^{2} + a^{2}) dx
⇒ ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2}) + a^{2} ∫1/√(x^{2} + a^{2}) dx  ∫(x^{2} + a^{2})/√(x^{2} + a^{2}) dx
⇒ ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2}) + a^{2} ∫1/√(x^{2} + a^{2}) dx  ∫√(x^{2} + a^{2}) dx
⇒ ∫√(x^{2} + a^{2}) dx + ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2}) + a^{2} ∫1/√(x^{2} + a^{2}) dx
⇒ 2 ∫√(x^{2} + a^{2}) dx = x√(x^{2} + a^{2}) + a^{2}[log x + √(x^{2} + a^{2}) + C]
⇒ ∫√(x^{2} + a^{2}) dx = (x/2)√(x^{2} + a^{2}) + (a^{2}/2) [log x + √(x^{2} + a^{2}) + C]
⇒ ∫√(x^{2} + a^{2}) dx = (x/2)√(x^{2} + a^{2}) + (a^{2}/2) log x + √(x^{2} + a^{2}) + K, where K = C(a^{2}/2)
Hence, the integral of root x square plus a square is given by, ∫√(x^{2} + a^{2}) dx = (x/2)√(x^{2} + a^{2}) + (a^{2}/2) log x + √(x^{2} + a^{2}) + K, where K is the integration constant. Similarly, we can determine the internal of root x square minus a square.
Integration of Root x Square Minus a Square
As we derived the integration of root x square plus a square, now we will determine the integration of root x square minus a square, that is, √(x^{2}  a^{2}). We will use the formula of integration ∫1/√(x^{2} + a^{2}) dx = log x + √(x^{2} + a^{2}) + C and integration by parts. Therefore, we have
∫√(x^{2}  a^{2}) dx = ∫√(x^{2}  a^{2}).1 dx
⇒ ∫√(x^{2}  a^{2}) dx = √(x^{2}  a^{2}) ∫dx  ∫[d(√(x^{2}  a^{2}))/dx ∫dx] dx
⇒ ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  ∫x^{2}/√(x^{2}  a^{2}) dx
⇒ ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  ∫(a^{2}  a^{2} + x^{2})/√(x^{2}  a^{2}) dx
⇒ ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  a^{2} ∫1/√(x^{2}  a^{2}) dx  ∫(x^{2}  a^{2})/√(x^{2}  a^{2}) dx
⇒ ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  a^{2} ∫1/√(x^{2}  a^{2}) dx  ∫√(x^{2}  a^{2}) dx
⇒ ∫√(x^{2}  a^{2}) dx + ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  a^{2} ∫1/√(x^{2}  a^{2}) dx
⇒ 2 ∫√(x^{2}  a^{2}) dx = x√(x^{2}  a^{2})  a^{2}[log x + √(x^{2}  a^{2}) + C]
⇒ ∫√(x^{2}  a^{2}) dx = (x/2)√(x^{2}  a^{2})  (a^{2}/2) [log x + √(x^{2}  a^{2}) + C]
⇒ ∫√(x^{2}  a^{2}) dx = (x/2)√(x^{2}  a^{2})  (a^{2}/2) log x + √(x^{2}  a^{2}) + K, where K = C(a^{2}/2)
Hence, the integral of root x square minus a square is given by, ∫√(x^{2}  a^{2}) dx = (x/2)√(x^{2}  a^{2})  (a^{2}/2) log x + √(x^{2}  a^{2}) + K, where K is the integration constant.
Important Notes on Integration of Root x
 The integration of root x is ∫√x dx = (2/3) x^{3/2} + C, where C is the integration constant.
 ∫√(x^{2}  a^{2}) dx = (x/2)√(x^{2}  a^{2})  (a^{2}/2) log x + √(x^{2}  a^{2}) + K
 ∫√(x^{2} + a^{2}) dx = (x/2)√(x^{2} + a^{2}) + (a^{2}/2) log x + √(x^{2} + a^{2}) + K
☛ Related Topics:
Integration of Root x Examples

Example 1: Evaluate the integration of root x plus one by root x, that is, √x + 1/√x.
Solution: To determine the integration of √x + 1/√x, we will use the formula of the integral of square root x. Also, we will use the formula ∫x^{n} dx = x^{n+1}/(n + 1) + C.
∫[√x + 1/√x] dx = ∫√x dx + ∫(1/√x) dx
= (2/3) x^{3/2} + 2√x + C
Answer: Hence, the integration of root x plus one by root x is (2/3) x^{3/2} + 2√x + C.

Example 2: Determine the integral of square root x minus one, that is, √(x  1).
Solution: To find the integral of √(x  1), we will use the substitution method.
Assume √(x  1) = u ⇒ 1/2√(x  1) dx = du ⇒ dx = 2u du
∫√(x  1) dx = ∫u 2u du
= ∫2u^{2} du
= 2u^{3}/3 + C
= (2/3) [√(x  1)]^{3} + C
= (2/3) (x  1)^{3/2} + C
Answer: The integral of square root x minus one is (2/3) (x  1)^{3/2} + C.
FAQs on Integration of Root x
What is Integration of Root x?
The integration of root x is (2/3) x^{3/2} + C, where C is the constant of integration. It can be determined by using the formula ∫x^{n} dx = x^{n+1}/(n + 1) + C.
How to Find Integral of Square Root x?
The integral of square root x can be found using the formula of integration ∫x^{n} dx = x^{n+1}/(n + 1) + C. In this formula, we can substitute n = 1/2 as root x can be written as √x = x^{1/2}.
What is the Integration of Root x Square Plus a Square?
The Integration of Root x Square Plus a Square is given by ∫√(x^{2} + a^{2}) dx = (x/2)√(x^{2} + a^{2}) + (a^{2}/2) log x + √(x^{2} + a^{2}) + K which is calculated using the integration by parts method of integration.
How to Find the Integral of Square Root x Square Minus a Square?
To find the integral of square root x square minus a square, we can use the method of integration by parts. The formula of this integration is ∫√(x^{2}  a^{2}) dx = (x/2)√(x^{2}  a^{2})  (a^{2}/2) log x + √(x^{2}  a^{2}) + K.
What is the Value of Definite Integration of Root x From 1 to 10?
The value of the definite integration of root x with limits from 1 to 10 is (2/3)(10^{3/2}  1).
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