如何在不使用双曲函数的前提下计算∫1/(1-7x²)² dx的积分？

Absolutely! You don’t need hyperbolic functions at all for this integral—we can tackle it using integration by parts and partial fraction decomposition, which are standard techniques covered in most calculus curricula. Let’s walk through the steps clearly:

Step 1: Use integration by parts via a derivative trick #

First, notice we can link our target integral to the derivative of a simpler function. Let’s compute the derivative of ( \frac{x}{1-7x^2} ):

d/dx [x/(1-7x²)] = [ (1)(1-7x²) - x*(-14x) ] / (1-7x²)²
= [1 -7x² +14x²]/(1-7x²)²
= (1 +7x²)/(1-7x²)²

Rewrite ( 1+7x² ) to relate it to the denominator: ( 1+7x² = 2 - (1-7x²) ). Substitute this into the derivative:

(2 - (1-7x²))/(1-7x²)² = 2/(1-7x²)² - 1/(1-7x²)

This gives us the key equation:

( \frac{d}{dx}\left( \frac{x}{1-7x^2} \right) = \frac{2}{(1-7x²⁾2} - \frac{1}{1-7x^2} )

Step 2: Rearrange to isolate the target integral #

Rearrange the equation to solve for the term we want to integrate:

2/(1-7x²)² = d/dx[x/(1-7x²)] + 1/(1-7x²)

Integrate both sides with respect to ( x ):

2 ∫ \frac{1}{(1-7x^2)^2} dx = \frac{x}{1-7x^2} + ∫ \frac{1}{1-7x^2} dx + C

Divide both sides by 2 to get our target integral on the left:

( ∫ \frac{1}{(1-7x²⁾2} dx = \frac{1}{2} \cdot \frac{x}{1-7x^2} + \frac{1}{2} ∫ \frac{1}{1-7x^2} dx + C )

Step 3: Compute the remaining integral with partial fractions #

Now we need to solve ( ∫ \frac{1}{1-7x^2} dx ). Factor the denominator: ( 1-7x^2 = (1 - \sqrt{7}x)(1 + \sqrt{7}x) ). Use partial fraction decomposition:

1/[(1 - √7x)(1 + √7x)] = A/(1 - √7x) + B/(1 + √7x)

Multiply through by the denominator:

1 = A(1 + √7x) + B(1 - √7x)

Solve for ( A ) and ( B ):

• Set ( x = 1/\sqrt{7} ): ( 1 = 2A ) → ( A = 1/2 )
• Set ( x = -1/\sqrt{7} ): ( 1 = 2B ) → ( B = 1/2 )

The integral now becomes:

∫ [1/(2(1 - √7x)) + 1/(2(1 + √7x))] dx
= (1/2) ∫ 1/(1 - √7x) dx + (1/2) ∫ 1/(1 + √7x) dx

Use substitution (let ( u = 1 - √7x ) for the first term, ( v = 1 + √7x ) for the second) to compute each part:

= (1/2)(-1/√7) ln|1 - √7x| + (1/2)(1/√7) ln|1 + √7x| + C
= (1/(2√7)) ln\left| \frac{1 + √7x}{1 - √7x} \right| + C

Step 4: Combine all terms #

Substitute this result back into our equation for the target integral:

∫ \frac{1}{(1-7x^2)^2} dx = \frac{x}{2(1-7x^2)} + \frac{1}{4√7} ln\left| \frac{1 + √7x}{1 - √7x} \right| + C

That’s our final answer—no hyperbolic functions involved, just core calculus tools!

内容的提问来源于stack exchange，提问作者Nehal Samee