# Evaluating $\int \arctan4t \, dt$

I am trying to evaluate $$\int \arctan4t \, dt$$

I make $u = \arctan4t$, $du = \frac{4}{1-4t}\,dt$

and $dv = dt$ and $v = t$

I then make it the form of $uv = \int v \, du$

$t\arctan4t - \int \frac{4t}{1-16t^2} \, dt$

Now to get the integral of $\int \frac{4t}{1-16t^2} \, dt$ I just pull out the 4 and use the identity I have memorized.

But I just now realised that this is not possible since I have a $16t^2$ so I have no idea how to advance from here.

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Now make the substitution $w = 1 - 16t^2$. – Henry T. Horton May 29 '12 at 22:25
@HenryT.Horton Be careful with variables. I would use something like $s$ since $v$ was used in the first round of integrating by parts (don't worry, I knew what you meant - others may not though.) – Joe May 29 '12 at 22:27
@JoeL.: I just noticed that and changed it. Thanks. – Henry T. Horton May 29 '12 at 22:28
Your $du$ is off. It should be $du={4\over 1+16t^2}$. Integration by parts then gives $t\arctan 4t-\int {4t\,dt\over 1+16t^2}$. To evaluate the last integral, use a regular substitution. $s= 1+16t^2$, $ds= 32t\,dt$. Then you have $t\arctan 4t-\int {4t\,dt\over 1+16t^2}=t\arctan 4t-{1\over8}\int{1\over s}\,ds=\cdots$ – David Mitra May 29 '12 at 22:30

Since $\frac{\mathrm{d}}{\mathrm{d}t}\arctan(4t)=\frac{4}{1+16t^2}$, integration by parts yields $$\int\color{red}{\arctan(4t)}\,\color{green}{\mathrm{d}t}=\color{green}{t}\color{red}{\arctan(4t)}-\int\color{green}{t}\color{red}{\frac{4}{1+16t^2}\mathrm{d}t}$$ Then let $u=1+16t^2$, so that $\frac{\mathrm{d}}{\mathrm{d}t}u=32t$, and we get \begin{align} \int\arctan(4t)\,\mathrm{d}t &=t\arctan(4t)-\int t\frac{4}{1+16t^2}\mathrm{d}t\\ &=t\arctan(4t)-\frac18\int\frac{\mathrm{d}u}{u}\\ &=t\arctan(4t)-\frac18\log(u)+C\\ &=t\arctan(4t)-\frac18\log(1+16t^2)+C \end{align}

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Let $s = 1+16t^2$. So $ds = 32t \;dt$ and thus $t \; dt = ds/32$. So we have that $$\int \dfrac{4t}{1+16t^2}dt = \dfrac{1}{8} \int \dfrac{ds}{s} = \dfrac{1}{8} \mbox{ln |s|} + C = \dfrac{1}{8} \mbox{ln |1+16t^2|} + C.$$

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More generally

$$\int \arctan x dx=x\arctan x-\frac{1}{2} \log\left( 1+x^2\right)+C$$

So that with $au=x$

$$\int \arctan au du =\frac{1}{a}\int\arctan x dx=\frac{x}{a}\arctan x-\frac{1}{2a} \log\left( 1+x^2\right)+C$$

$$\int \arctan au du ={u}\arctan au-\frac{1}{2a} \log\left( 1+a^2u^2\right)+C$$

The first result is obtain quite striaghtforwarly from integrating by parts with $dx=du$ and $\arctan x =v$.

\eqalign{ & \int {\arctan } xdx = x\arctan x - \int {\frac{{xdx}}{{1 + {x^2}}}} \cr & \int {\arctan } xdx = x\arctan x - \frac{1}{2}\int {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} \cr & \int {\arctan } xdx = x\arctan x - \frac{1}{2}\log \left( {1 + {x^2}} \right) + C \cr}

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As noted by David in the comments, your $du$ is off. Your $du$ should be $\frac{4 \, dt}{1+16t^2}.$

So, we have:

$$t\arctan(4t) - \int \frac{4t \, dt}{1+16t^2}$$

Then, you can let $s = 1 + 16t^2$ and $ds = 32t \, dt$.

$$t\arctan(4t) - \frac{1}{8} \int \frac{1}{s} \, ds$$

$$t\arctan(4t) - \frac{1}{8} \ln {(1+16t^2)} + C$$

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I guess you're right my mistake. – Eugene May 29 '12 at 22:55