Integral Of 1 Sqrt X 2 4

Integral of (\displaystyle \frac{1}{\sqrt{x},(x^{2}+4)}): A Step‑by‑Step Guide

Introduction

When you first encounter the integral
[ \int \frac{dx}{\sqrt{x},(x^{2}+4)}, ] it can appear intimidating because it combines a radical in the denominator with a quartic polynomial. Still, by applying a suitable substitution and a little algebraic manipulation, the problem becomes approachable. This article walks through the solution in detail, explains the underlying ideas, and provides additional insights that can help you tackle similar integrals.

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1. Recognizing the Structure

The integrand contains two distinct parts:

1. A radical factor: (\sqrt{x}) in the denominator suggests a substitution that removes the square root.
2. A quartic factor: (x^{2}+4) remains after the substitution.

A common strategy is to let (x) be a perfect square, i.Here's the thing — e. , set (x = t^{2}). This eliminates the square root and converts the integral into a rational function of (t), which is often easier to handle.

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2. Substitution (x = t^{2})

Let
[ x = t^{2} \quad \Longrightarrow \quad dx = 2t,dt. ]

Also, [ \sqrt{x} = \sqrt{t^{2}} = t \quad (\text{since } t \ge 0 \text{ when } x \ge 0). ]

Substituting into the integral:

[ \int \frac{dx}{\sqrt{x},(x^{2}+4)} = \int \frac{2t,dt}{t,(t^{4}+4)} = \int \frac{2,dt}{t^{4}+4}. ]

Now the problem reduces to integrating the rational function (\displaystyle \frac{2}{t^{4}+4}).

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3. Factoring the Denominator

The quartic (t^{4}+4) can be factored over the reals (and complex numbers). Notice that

[ t^{4}+4 = t^{4}+4t^{2}+4 - 4t^{2} = (t^{2}+2)^{2} - (2t)^{2} = (t^{2}+2-2t)(t^{2}+2+2t). ]

Thus, [ t^{4}+4 = (t^{2}-2t+2)(t^{2}+2t+2). ]

Both quadratic factors have positive discriminants? Let's check:

• For (t^{2}-2t+2): discriminant (D = (-2)^{2} - 4\cdot1\cdot2 = 4-8 = -4 < 0).
• For (t^{2}+2t+2): discriminant (D = 4-8 = -4 < 0).

So both quadratics are irreducible over the reals, but we can still perform partial fraction decomposition using them.

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4. Partial Fraction Decomposition

We seek constants (A,B,C,D) such that

[ \frac{2}{(t^{2}-2t+2)(t^{2}+2t+2)} = \frac{At+B}{t^{2}-2t+2} + \frac{Ct+D}{t^{2}+2t+2}. ]

Multiplying both sides by the denominator gives

[ 2 = (At+B)(t^{2}+2t+2) + (Ct+D)(t^{2}-2t+2). ]

Expanding and collecting like powers of (t):

[ \begin{aligned} 2 &= (At+B)(t^{2}+2t+2) + (Ct+D)(t^{2}-2t+2) \ &= At^{3} + 2At^{2} + 2At + Bt^{2} + 2Bt + 2B \ &\quad + Ct^{3} - 2Ct^{2} + 2Ct + Dt^{2} - 2Dt + 2D. \end{aligned} ]

Group terms:

• (t^{3}): (A + C)
• (t^{2}): (2A + B - 2C + D)
• (t): (2A + 2B + 2C - 2D)
• Constant: (2B + 2D)

Set each coefficient equal to the corresponding coefficient on the left side (which is (0) for (t^{3}, t^{2}, t) and (2) for the constant):

[ \begin{cases} A + C = 0, \ 2A + B - 2C + D = 0, \ 2A + 2B + 2C - 2D = 0, \ 2B + 2D = 2. \end{cases} ]

From the first equation: (C = -A).

From the fourth: (B + D = 1).

Plugging (C = -A) into the second:

[ 2A + B + 2A + D = 0 \quad \Rightarrow \quad 4A + (B + D) = 0. ]

But (B + D = 1), so (4A + 1 = 0 \Rightarrow A = -\tfrac{1}{4}).

Thus (C = \tfrac{1}{4}).

Now use (B + D = 1) and the third equation:

[ 2A + 2B + 2C - 2D = 0 \quad \Rightarrow \quad 2(-\tfrac{1}{4}) + 2B + 2(\tfrac{1}{4}) - 2D = 0. ]

The (-\tfrac{1}{2}) and (\tfrac{1}{2}) cancel, leaving (2B - 2D = 0 \Rightarrow B = D).

Since (B + D = 1) and (B = D), we get (B = D = \tfrac{1}{2}).

So the decomposition is:

[ \frac{2}{t^{4}+4} = \frac{-\tfrac{1}{4}t + \tfrac{1}{2}}{t^{2}-2t+2}

• \frac{\tfrac{1}{4}t + \tfrac{1}{2}}{t^{2}+2t+2}. ]

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5. Integrating Each Term

We now integrate term by term.

5.1. Integral of (\displaystyle \frac{-\tfrac{1}{4}t + \tfrac{1}{2}}{t^{2}-2t+2})

Complete the square in the denominator:

[ t^{2} - 2t + 2 = (t-1)^{2} + 1. ]

Let (u = t-1). Then (du = dt) and (t = u+1). Substitute:

[ \int \frac{-\tfrac{1}{4}(u+1) + \tfrac{1}{2}}{u^{2}+1},du = \int \frac{-\tfrac{1}{4}u - \tfrac{1}{4} + \tfrac{1}{2}}{u^{2}+1},du = \int \frac{-\tfrac{1}{4}u + \tfrac{1}{4}}{u^{2}+1},du. ]

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Split into two integrals:

[ -\frac{1}{4}\int \frac{u}{u^{2}+1},du + \frac{1}{4}\int \frac{1}{u^{2}+1},du. ]

These are standard:

• (\displaystyle \int \frac{u}{u^{2}+1},du = \frac{1}{2}\ln(u^{2}+1)).
• (\displaystyle \int \frac{1}{u^{2}+1},du = \arctan u).

Thus,

[ -\frac{1}{4}\cdot\frac{1}{2}\ln(u^{2}+1) + \frac{1}{4}\arctan u = -\frac{1}{8}\ln(u^{2}+1) + \frac{1}{4}\arctan u. ]

Re‑express (u = t-1):

[ -\frac{1}{8}\ln!\bigl((t-1)^{2}+1\bigr) + \frac{1}{4}\arctan(t-1). ]

5.2. Integral of (\displaystyle \frac{\tfrac{1}{4}t + \tfrac{1}{2}}{t^{2}+2t+2})

Similarly, complete the square:

[ t^{2} + 2t + 2 = (t+1)^{2} + 1. ]

Let (v = t+1). Then (dv = dt) and (t = v-1). Substitute:

[ \int \frac{\tfrac{1}{4}(v-1) + \tfrac{1}{2}}{v^{2}+1},dv = \int \frac{\tfrac{1}{4}v - \tfrac{1}{4} + \tfrac{1}{2}}{v^{2}+1},dv = \int \frac{\tfrac{1}{4}v + \tfrac{1}{4}}{v^{2}+1},dv. ]

Again split:

[ \frac{1}{4}\int \frac{v}{v^{2}+1},dv + \frac{1}{4}\int \frac{1}{v^{2}+1},dv = \frac{1}{4}\cdot\frac{1}{2}\ln(v^{2}+1) + \frac{1}{4}\arctan v. ]

Thus,

[ \frac{1}{8}\ln!\bigl((t+1)^{2}+1\bigr) + \frac{1}{4}\arctan(t+1). ]

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6. Combining the Results

Add the two antiderivatives:

[ \begin{aligned} \int \frac{2}{t^{4}+4},dt &= \Bigl[-\frac{1}{8}\ln!Even so, \bigl((t-1)^{2}+1\bigr) + \frac{1}{4}\arctan(t-1)\Bigr] \ &\quad + \Bigl[\frac{1}{8}\ln! \bigl((t+1)^{2}+1\bigr) + \frac{1}{4}\arctan(t+1)\Bigr] + C.

The logarithmic terms partially cancel:

[ -\frac{1}{8}\ln!\bigl((t-1)^{2}+1\bigr) + \frac{1}{8}\ln!\bigl((t+1)^{2}+1\bigr) = \frac{1}{8}\ln!\frac{(t+1)^{2}+1}{(t-1)^{2}+1}. ]

So the antiderivative simplifies to

[ \boxed{ \int \frac{2}{t^{4}+4},dt = \frac{1}{8}\ln!\frac{(t+1)^{2}+1}{(t-1)^{2}+1}

• \frac{1}{4}\bigl[\arctan(t+1) + \arctan(t-1)\bigr] + C. } ]

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7. Returning to the Original Variable

Recall that (t = \sqrt{x}). Substitute back:

[ \boxed{ \int \frac{dx}{\sqrt{x},(x^{2}+4)} = \frac{1}{8}\ln!\frac{(\sqrt{x}+1)^{2}+1}{(\sqrt{x}-1)^{2}+1}

• \frac{1}{4}\Bigl[\arctan(\sqrt{x}+1) + \arctan(\sqrt{x}-1)\Bigr] + C. } ]

This expression is the exact antiderivative. For practical purposes, you might want to simplify the logarithm’s argument:

[ (\sqrt{x}\pm 1)^{2}+1 = x \pm 2\sqrt{x} + 2. ]

Thus,

[ \boxed{ \int \frac{dx}{\sqrt{x},(x^{2}+4)} = \frac{1}{8}\ln!\frac{x+2\sqrt{x}+2}{,x-2\sqrt{x}+2,}

• \frac{1}{4}\Bigl[\arctan(\sqrt{x}+1) + \arctan(\sqrt{x}-1)\Bigr] + C. } ]

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8. Key Takeaways

• Substitution first: Replacing (x) with a square ((t^{2})) removes the radical and simplifies the integrand to a rational function.
• Factor the denominator: Recognizing that (t^{4}+4) factors into two irreducible quadratics allows partial fraction decomposition.
• Complete the square: This technique transforms each quadratic denominator into the standard form (u^{2}+1), enabling the use of logarithm and arctangent integrals.
• Combine logarithms: When two logarithmic terms appear with opposite signs, they can often be merged into a single logarithm of a ratio.
• Back‑substitute carefully: Always revert to the original variable, simplifying expressions where possible.

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FAQ

Question	Answer
Why choose (x = t^{2}) instead of (x = u^{2}) or (x = t^{4})?	Using (x = t^{2}) directly cancels the (\sqrt{x}) factor. Other substitutions either leave a radical or complicate the polynomial.
Can we integrate (\frac{1}{\sqrt{x}(x^{2}+4)}) using trigonometric substitution?	Yes, but it is more involved. Which means the algebraic substitution shown here is more straightforward for this integrand. And
**Is the result unique? Plus, **	Any antiderivative can differ by a constant. And the form presented is one convenient representation. Consider this:
**What if (x) is negative? Here's the thing — **	The integral is defined for (x>0) because of (\sqrt{x}). For negative (x), a complex substitution would be required.

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Conclusion

The integral of (\displaystyle \frac{1}{\sqrt{x},(x^{2}+4)}) is a beautiful example of how algebraic manipulation, partial fractions, and trigonometric identities come together. Think about it: by systematically transforming the integrand, we arrive at a compact closed‑form expression involving only elementary functions. Mastering these techniques equips you to tackle a wide array of integrals that blend radicals with polynomials, opening the door to deeper exploration in calculus and beyond.