# What is `int1/(1+t^2) (veci +tvecj + t^2veck) dt`?

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I have to apologize, the answer was cut off. The full answer is as follows:

`veci(tan^-1t + C_x) + vecj(lnsqrt(1+t^2) + C_y) + veck(t-tan^-1t+C_z)`

To integrate that function, we need to separated out the integral into a sum of three integrals by first simplifying the expression in the interior and then distributing the integration:

`=int 1/(1+t^2)veci + t/(1+t^2)vecj + t^2/(1+t^2)veck dt`

`=int (vecidt)/(1+t^2) + int (tvecj)/(1+t^2) dt + int (t^2veck)/(1+t^2) dt`

Now, we can recognize that the vector terms are constants and can be removed from the integrals:

`=veciint dt/(1+t^2) + vecjint t/(1+t^2) dt + veckint t^2/(1+t^2) dt`

Now, we need to integrate each expression. We're going to use `C_(x),C_y,C_z` as constants of integration in each of the basis vector directions.

The first integral, based on trigonometric identities can be found to be (see first link for an integral table of rational functions):

`veci int dt/(1+t^2) = veci(tan^-1t +C_(x))`

The second integral can be better solved by letting `u = 1+t^2` giving, therefore, `du = 2t dt`. We can now substitute `(du)/2` for `tdt` to give:

`vecj int(tdt)/(1+t^2)=vecj int 1/2*(du)/u`

We can now evaluate the integral:

`=vecj/2 (lnu) = vecj/2*ln(1+t^2) = vecj(lnsqrt(1+t^2) + C_(y))`

Finally, we will evaluate the third integral:

`veck int (t^2dt)/(1+t^2)`

Let's go ahead and make some trigonometric substitutions:

`t = tantheta`

This means when we substitute for dt:

`dt = sec^2theta d theta`

We now get the following:

`veckint(t^2dt)/(1+t^2) =veckint(tan^2thetasec^2theta d theta)/(1+tan^2theta)`

This wil look a lot more manageable when we realize that `1+tan^2theta = sec^2theta` :

`=veck int (tan^2thetasec^2theta d theta)/(sec^2theta)`

Now, we can cancel out the secants and end up with a manageable integral.

`=veck int tan^2 theta d theta`

Using the integral table (for `int tan^n (ax) dx`) from the second link below, we can see this integral can be evaluated as

`=veck (tantheta - theta + C_z)`

Now, when we substitute t back in for `tantheta`:

`=veck(t - tan^-1t + C_z)`

Now, we have our completely-evaluated integral:

`int(veci + tvecj + t^2veck)/(1+t^2) dt = veci(tan^-1t +C_x) + vecj(lnsqrt(1+t^2) + C_y)+veck(t-tan^-1t + C_z)`

I hope you were able to follow this! Good luck!

**Sources:**