Integration of Root(a^2+x^2) – iMath

Integral of sqrt(1-2x)
Integral of sqrt(1-2x)

The integration of the square root of a2+x2 is given as follows:

$\int \sqrt{a^2+x^2} dx$ $=\dfrac{x}{2} \sqrt{a^2+x^2}$ $+\dfrac{a^2}{2}\log |x+\sqrt{a^2+x^2}|+C$, where $C$ is an integration constant.

In this post, we will find the integral of root(a2+x2). Let’s learn how to integrate square root of a2+x2.

Integration of $\sqrt{a^2+x^2}$

Question: What is the integration of square root of $a^2+x^2$?

Answer: The integration of root of $a^2+x^2$ is $\dfrac{x}{2} \sqrt{a^2+x^2}$ $+\dfrac{a^2}{2}\log |x+\sqrt{a^2+x^2}|+C$.

Explanation:

Let $I = \int \sqrt{a^2+x^2} dx$ $\cdots (I)$

To find the integration of the square root of a2+x2, we will use the integration by parts formula. The integration by parts formula says that if f(x), g(x) are two functions then the integration of the product f(x)g(x) is as follows:

$\int f(x)g(x) dx$ $=f \int g dx$ $-\int [\dfrac{df}{dx} \int g dx]dx$ $\cdots (\star)$

In this formula $(\star)$, we put $f(x)=\sqrt{a^2+x^2}$ and $g(x)=1$.

First, we calculate $\int g(x) dx$ $=\int 1 dx =x$

and

$\dfrac{d}{dx}(f(x))$ $=\dfrac{d}{dx}(\sqrt{a^2+x^2})$ $=\dfrac{d}{dx}((a^2+x^2)^{1/2})$

$=\dfrac{1}{2}(a^2+x^2)^{1/2-1}\cdot \dfrac{d}{dx}(a^2+x^2)$ by the chain rule and power rule of integration.

$=\dfrac{1}{2}(a^2+x^2)^{-1/2}\cdot 2x$

$=\dfrac{x}{\sqrt{a^2+x^2}}$

Now putting $f(x)=\sqrt{a^2+x^2}$ and $g(x)=1$ in the formula $(\star)$, the integral of root(a^2+x^2) is

$I=\int \sqrt{a^2+x^2} dx = \int \sqrt{a^2+x^2} \cdot 1 dx$

$=\sqrt{a^2+x^2} \cdot x$ $-\int \dfrac{x}{\sqrt{a^2+x^2}} \cdot x dx$

$=x\sqrt{a^2+x^2}$ $-\int \dfrac{a^2+x^2-a^2}{\sqrt{a^2+x^2}}$

$=x\sqrt{a^2+x^2}-\int \sqrt{a^2+x^2} dx$ $+a^2 \int \dfrac{dx}{a^2+x^2}$

Therefore,

$I=x\sqrt{a^2+x^2}-I$ $+a^2 \int \dfrac{dx}{a^2+x^2}$

⇒ $2I = x\sqrt{a^2+x^2}$ $+a^2 \int \dfrac{dx}{a^2+x^2}$

⇒ $2I=x\sqrt{a^2+x^2}$ $+a^2 \log |x+\sqrt{a^2+x^2}|+C’$

So the integration of root(a^2+x^2) is equal to $I= \int \sqrt{a^2+x^2} dx$ $=\dfrac{x}{2} \sqrt{a^2+x^2}$ $+\dfrac{a^2}{2}\log |x+\sqrt{a^2+x^2}|+C$, where $C=\dfrac{C’}{2}$ is an integration constant.

Also Read:

Integration of log(sinx) from 0 to pi/2

Derivative & integration of 1/root(x)

Question-Answer on integration of $\sqrt{a^2+x^2}$

Question: Find the integral $\int \sqrt{25+x^2} dx$.

Answer:

Put $a=5$ in the above formula. Then the integration of $\sqrt{25+x^2}$ will be equal to

$\int \sqrt{25+x^2} dx$

= $\int \sqrt{5^2+x^2} dx$

= $\dfrac{x}{2} \sqrt{25+x^2}$ $+\dfrac{25}{2}\log |x+\sqrt{25+x^2}|+C$

Here $C$ is the constant of integrations.

FAQs

Q1: What is the integration of square root of a2+x2?

Answer: The integration of square root of a2+x2 is equal to ∫√(a2+x2) dx =x√(a2+x2)/2 + a2/2 log |x+√(a2+x2)|+C where C is an integration constant.

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