Prove that

$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$

My try:

$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$

$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$

$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$

Prove $(1)$ using an alternative method other than substitution method.

  • 2
    $\begingroup$ Generate an accurate drawing of the curve on a pair of axes (1 unit = 1cm), and 3d print a cylinder whose cross-section is the given region, having height 1 cm. Fill it with water, and then place the water into a cup and place it on a balance beam. Subtract the mass of the cup and, if the result is pretty close to 1/2 of a gram, you got it. $\endgroup$
    – David P
    Commented Jan 3, 2017 at 19:39
  • $\begingroup$ Have you tried partial fractions? $\qquad$ $\endgroup$ Commented Jan 3, 2017 at 19:48
  • $\begingroup$ @DavidP : While you're at it, why not tell him to get into his bathtub? That worked for Archimedes. $\endgroup$ Commented Jan 3, 2017 at 19:50
  • $\begingroup$ in fact very time I took a bath I got a good idea of formula to post on MSE. Not lying. $\endgroup$ Commented Jan 3, 2017 at 19:54
  • $\begingroup$ Hint: Adding $4-4=0$ in the numerator splits the integral into two trivial parts $\endgroup$
    – tired
    Commented Jan 4, 2017 at 1:19

4 Answers 4


Set $\frac{3-\sqrt{5}x}{1+\sqrt{5}x}=u$, we have $\frac{-4\sqrt{5}} {(1+\sqrt{5}x)^2} dx=du$ thus $$\int_{0}^{1} {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx=\frac{1}{-4\sqrt{5}}\int_{3}^{\sqrt{5}-2}u\,du=\frac 12$$

  • 1
    $\begingroup$ Easy lemon squeezy $\endgroup$ Commented Jan 3, 2017 at 20:25
  • $\begingroup$ This is actually a substitution method while you asked about one not using it. $\endgroup$
    – Blencer
    Commented Jan 7, 2017 at 15:53

Using integration by parts : $$\begin{align}\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx &= \int_0^1 \left( - \frac{1}{2\sqrt{5} (1+\sqrt{5}x)^2} \right)'(3-\sqrt{5}x) \, dx \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2}\right]_0^1 - \frac{1}{2} \int_0^1 \frac{dx}{(1+\sqrt{5}x)^2} \\[8pt] &= \left[ \frac{\sqrt{5}x - 3}{2\sqrt{5}(1+\sqrt{5}x)^2} + \frac{1}{10x + 2\sqrt{5}}\right]_0^1 \\[8pt] &= \frac{1}{2} \end{align} $$

  • $\begingroup$ Good catch! (+1) $\endgroup$ Commented Jan 3, 2017 at 19:36

$$ {3-\sqrt{5}x\over (1+\sqrt 5 \, x)^3} = \frac A {1+\sqrt 5\,x} + \frac B {(1+\sqrt 5\,x)^2} + \frac C {(1+\sqrt 5\,x)^3} $$ Find $A$, $B$, and $C$ and then integrate each term separately.


Set $x=\frac{\sinh^2u}{\sqrt{5}}$ and $dx=\frac{2\sinh u\cosh u}{\sqrt{5}}$

So $$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx=\int \frac{6\sinh u\cosh u}{\sqrt{5}\cosh^6 u}du -\int\frac{2\sinh^3u \cosh u}{\sqrt{5}\cosh^6 u} \, du = \int \frac{6\sinh u}{\sqrt{5}\cosh^5 u} \, du -\int \frac{2(1+\cosh^2 u ) \sinh u}{\sqrt{5}\cosh^5 u} \, du$$

And you can calculate these integrals using the substitution $t=\cosh u$ and $dt=\sinh u \, du$

Another way using partial fractions:

$$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} dx= \int \frac{20\sqrt{5}}{(5x+\sqrt{5})^3}- \frac{5}{(5x+\sqrt{5})^2}dx $$


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