Question

how do you compute this indefinite integral ?
∫(5x^2)/(x^2+1)dx

Answer 1

1)you can take that 5 out
2)in num,,you have only x^2 ,, write it as (x^2 + 1) -1
3)just a simple simplification and you get 1 - (1/ (x^2 +1)) inside the integral
rest should be easier..

Answer 2

You should break the numerator first I think..

Answer 3

do u have options

Answer 4

\[5\int\limits_{}^{}\frac{x^2 +1 - 1}{x^2 + 1}dx = 5\int\limits_{}^{}dx - 5\int\limits_{}^{}\frac{dx}{x^2 + 1}\]

Answer 5

For solving later one use the formula:

\[\int\limits_{}^{}\frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C\]

Answer 6

thanks a lot, so when there is x to the power of something in the numerator you always have to simplify the numerator right?

Answer 7

depends on the denom also..you'll understand better on practicing more..

Answer 8

how come x+1 didnt get cancel out ?

Answer 9

No, no it is not necessary..

Answer 10

x^2 +1 *

Answer 11

Where what you are talking about??

Answer 12

See, I show you..