I am trying to solve the following equation:


As far as I can remember I have to move x to left and t to right and form an integration. What should I do? Does deviding both side by $(t+4)$ helps? what will be next?!


ok as mrf suggested I came to this:

$$\int \frac 1{1+x^2}\,dx = 4\int\frac 1{t+4}\,dt$$

This will be:

$$\tan^{-1}(x)+C == $$

I am not sure about right side...any tips?!

  • 1
    $\begingroup$ Separating variables sounds like a good idea. Write the equation as $$\frac{dx}{1+x^2} = \frac{4\,dt}{t+4}$$ and integrate both sides. $\endgroup$ – mrf May 2 '12 at 19:15
  • $\begingroup$ Thanks, can you check my update please?! $\endgroup$ – Sean87 May 2 '12 at 19:26
  • 1
    $\begingroup$ What's the derivative of $\ln |t|$? How about $\ln |t+4|$? $\endgroup$ – mrf May 2 '12 at 19:27
  • $\begingroup$ Thats true only if numinator is 1 ? $\endgroup$ – Sean87 May 2 '12 at 19:32
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    $\begingroup$ Remember... $$\frac{d}{dx} 4\log |t+4| = \frac{4}{t+4}$$ $\endgroup$ – Argon May 2 '12 at 19:42

As you rearranged it in your post, we have

$$\int \frac{dx}{1+x^2} = 4\int \frac{dt}{t+4}$$

As you discovered, the LHS is $\arctan$. To find the value of the RHS, we may say $t+4=u$ so $du=dt$. Thus the RHS can be rewritten as

$$4\int \frac{du}{u} = 4\log |u|+C_2$$

Because we want this integral to be in respect to $t$, we have $u=t+4$, so

$$\arctan x +C_1= 4\log |t+4|+C_2$$

To confirm that the integral we solved is indeed correct, you can simply differentiate $4\log |t+4|+C_2$.


$x= \tan (4\log|t+4|+C_2-C_1)$


$t = \exp (\frac{1}{4}(\arctan x+C_1-C_2))-4$

  • $\begingroup$ Can you please explain how at the end you went from x to t? $\endgroup$ – Sean87 May 2 '12 at 23:20
  • $\begingroup$ @Sean87 What do you mean? Are you referring to when I solved for $x$ and $t$? $\endgroup$ – Argon May 2 '12 at 23:46
  • $\begingroup$ Yeah, why x = tan but t= exp ? $\endgroup$ – Sean87 May 3 '12 at 9:57
  • $\begingroup$ @Sean87 I am simply isolating $x$ and $t$ from $$\arctan x+C_1=4\log|t+4|+C_2 \implies \arctan x=4\log|t+4|+C_2-C_1 \implies x= \tan (4\log|t+4|+C_2-C_1)$$ Similarly, $$\arctan x+C_1=4\log|t+4|+C_2 \implies \arctan x+C_1-C_2=4\log |t+4| \implies \frac{1}{4}(\arctan x+C_1-C_2)=\log |t+4|$$ Assuming $|t+4|>0$ we have $$\frac{1}{4}(\arctan x+C_1-C_2)=\log |t+4| \implies \exp (\frac{1}{4}(\arctan x+C_1-C_2)) = |t+4|=t+4 \implies \exp (\frac{1}{4}(\arctan x+C_1-C_2))-4=t$$ $\endgroup$ – Argon May 3 '12 at 19:47

Here are some proceeding steps:

  • Seperating the variables you get $\displaystyle \frac{dx}{1+x^2} = 4 \cdot \frac{dt}{t+4}$

  • Integrating both sides you have: $\tan^{-1}(x) + C_{1} = 4 \cdot \log|t+4| + C_{2}$.


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