Question

practice

Answer 1

\(\LARGE\color{blue}{\int\limits_{ }^{ } \frac{3t}{(t^2+1)^2}~~dt }\)

\(\LARGE\color{blue}{3\int\limits_{ }^{ } \frac{t}{(t^2+1)^2} ~~dt}\)

subbing u for t²+1 and du= 2t dt

\(\LARGE\color{blue}{3\int\limits_{ }^{ } \frac{1}{u^2} ~~du}\)

\(\LARGE\color{blue}{= -\frac{3}{2u} }\)

\(\LARGE\color{blue}{= -\frac{3}{2(t^2+1)} }\)

\(\LARGE\color{blue}{= -\frac{3}{2t^2+2} }\)

Answer 2

Left out +C... well doesn't matter really, my teacher would take off.

Answer 3

When you said du=2tdt you should say du/2=tdt so that you don't lose that 1/2 factor in there.

Answer 4

Yes, I couldn't process what you told me, but I realize now ..

Answer 5

I'll draw them.

Answer 6

btw, sometimes having a few known integrals will help you rather a lot (there are usually tables in the front and/or back of the book) for instance, knowing the defined integral:
\[\int \frac{x}{(x^2+1)^2}dx=\frac1{2x^2+2}+C\] would make that first one a joke (also this is one that pops up rather often as parts of other functions)

Answer 7

u=1-x^4
du=-4x^3 dx

skipped a step I think...