Question

How to Calculate the integral of 15600/ (t^2-24t+160) ?

Answer 1

\[\int\limits\limits_{Ind.}^{Ind.}\frac{ 15600 }{ t ^{2}-24t+160 }dt=15600\int\limits\limits_{ind.}^{ind.}\frac{ 1 }{ t ^{2}-24t+160 }dt=-3900\tan^{-1} (3-\frac{ t }{ 4 })\]

Answer 2

Which of the following is not a step used to solve multi-step equations?
A.
identify values of variables in the equation from the written word problem
B.
finish by getting all variables and constants on the same side of the equation
C.
plug in known variables from the word problem into the equation
D.
translate the written phrases into an algebraic equation

Answer 3

\[\int\limits \frac{ 1 }{ t^2-24t+144+16 }dt=\int\limits \frac{ 1 }{\left( t-12 \right)^2+4^2 }dt=?\]
\[\int\limits \frac{ 1 }{x^2+a^2 }dx=\frac{ 1 }{a }\tan^{-1} \frac{ x }{ a }\]

Answer 4

I am confused about the development of the equation

Answer 5

Where did the 15600 went and how come there is a tan?

Answer 6

multiply by 15600,i have not completed the problem,only hint.

Answer 7

\[I=\int\limits \frac{ 1 }{ x^2+a^2 }dx,put~ x=a \tan \theta,dx=a~ \sec ^2 \theta~ d \theta\]
\[I=\int\limits \frac{ a \sec ^2 \theta d \theta }{a^2\tan ^2\theta+a^2 }=\frac{ 1 }{ a }\int\limits \frac{ \sec ^2\theta d \theta }{ \tan ^2 \theta+1 }\]
\[I=\frac{ \frac{ }{1 } }{a }\int\limits\limits \frac{ \sec ^2 \theta d \theta }{\sec ^2 \theta }=\frac{ 1 }{a } \int\limits\limits d \theta =\frac{ 1 }{ a } \theta=\frac{ 1 }{a }\tan^{-1} \frac{ x }{ a } \]

Answer 8

\[\frac{ x }{a }=\tan \theta ,\theta=\tan^{-1} \frac{ x }{a },\sec ^2 \theta-\tan ^2 \theta=1,\sec ^2\theta =1+\tan ^2 \theta \]

Answer 9

I get it, thanks

Answer 10

yw