# How to Evaluate $\int_1^4 (\frac{1}{2t}+i)^2 dt$

Question:

$$\int \limits _1^4\left (\frac{1}{2t}+i \right )^2\,dt.$$

How can I solve this? I believe its an indefinite integral and I can probably expand it by using $$(a+b)^2=a^2+2ab+b^2$$ to get something like$$\int \frac{1}{4t^2}+\frac{2i}{2t}+i^2\,dt$$but I'm not even sure if that's right.

• That $i$ it's a bad notation here. The $\int_{1}^{4}f(t)\,{\rm d}t$ is a definite integral and yes, you can use $(a+b)^{2}=a^{2}+2ab+b^{2}$. Then use $\int \frac{1}{t}\, {\rm d}t=\ln|t|+c$ and $\int t^{n}dt=\frac{t^{n+1}}{n+1}+C$. May 17 at 16:32

Since $$\left(\frac1{2t}+i\right)^2=\frac1{4t^2}-1+\frac it,$$ you have \begin{align*} \int_1^4\left(\frac1{2t}+i\right)^2\,dt&=\int_1^4\frac1{4t^2}-1\,dt+\left(\int_1^4\frac1t\,dt\right)i\\ &=-\frac{45}{16}+i\log(4). \end{align*}
We have \begin{align*}\int \limits _1^4\left (\frac{1}{2t}+i\right )^2\,dt & =\int \limits _1^4\left (\frac{i}{t}+\frac{1}{4t^2}-1\right )\,dt \\ & =i\int \limits _1^4\frac{1}{t}\,dt+\frac{1}{4}\int \limits _1^4\frac{1}{t^2}\,dt-\int \limits _1^41\,dt \\ & =\left .i\ln |t|-t-\frac{1}{4t}\right |_1^4 \\ & =\frac{16i\ln 4-65}{16}+\frac{5}{4} \\ & =\frac{16i\ln 4-45}{16}. \end{align*}
• If $i^{2}=-1$ so what's the problem? There's not problem. However, I agree that to use $i$ is a bad notation here. May 17 at 16:34
• @EventHorizon edit: I automatically assumed that $i^2 = -1$ anyways to calculate it... May 17 at 16:36