Question

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Answer 1

\[\frac{ d }{ dx }(\frac{ x }{ x+4 })=\frac{ 4 }{ (x+4)^2 }.Hence,find~\int\limits \frac{ 4 }{ (x+4)^{2}~ }dx\]

Answer 2

@ganeshie8

Answer 3

What do youknow about the fundamental thoerem of calculus 1 ?

Answer 4

\[F(x)=\int\limits_{a}^{x}f(t)~dt\]

Answer 5

what do you mean ?

Answer 6

i'm not very sure
can u pls explain to me?

Answer 7

Fundamental theorem of calculus part 1 says this :

If \(F(x)=\int\limits_{a}^{x}\color{blue}{f(t)}~dt\), then :
\[\dfrac{d}{dx}F(x) =\color{blue}{ f(x)}\]

Answer 8

Also, we can show below :

If \(\dfrac{d}{dx}F(x) =\color{blue}{ f(x)}\), then :

\(\int \color{blue}{ f(x)}\,dx =F(x) + C\)

Answer 9

Read the given problem again

Answer 10

We're given \(\dfrac{d}{dx}\dfrac{x}{x+4} =\color{blue}{ \dfrac{4}{(x+4)^2}}\), and we need to evaluate

\(\int \color{blue}{\dfrac{4}{(x+4)^2}}dx=?\)

Answer 11

is it \[\int\limits\limits \frac{ 4 }{ (x+4)^2 }~dx=\frac{ x }{ x+4 }\]

Answer 12

Very close! just add +C

Answer 13

Thank you @ganeshie8

Answer 14

yw