Question

Question #5e4be

Answer 1

`1/(2sqrt9.8)lnabs((sqrt9.8+x)/(sqrt9.8-x))+C`

Explanation:

Note that:

`1/(9.8-x^2)=1/((sqrt9.8+x)(sqrt9.8-x))`

`color(white)(1/(9.8-x^2))=A/(sqrt9.8+x)+B/(sqrt9.8-x)`

So:

`1=A(sqrt9.8-x)+B(sqrt9.8+x)`

Letting `x=sqrt9.8`:

`1=B(2sqrt9.8)" "=>" "B=1/(2sqrt9.8)`

Letting `x=-sqrt9.8`:

`1=A(2sqrt9.8)" "=>" "A=1/(2sqrt9.8)`

Thus:

`1/(9.8-x^2)=1/(2sqrt9.8)(1/(sqrt9.8+x))+1/(2sqrt9.8)(1/(sqrt9.8-x))`

So:

`intdx/(9.8-x^2)=1/(2sqrt9.8)intdx/(sqrt9.8+x)+1/(2sqrt9.8)intdx/(sqrt9.8-x)`

`color(white)(intdx/(9.8-x^2))=1/(2sqrt9.8)lnabs(sqrt9.8+x)-1/(2sqrt9.8)lnabs(sqrt9.8-x)`

`color(white)(intdx/(9.8-x^2))=1/(2sqrt9.8)lnabs((sqrt9.8+x)/(sqrt9.8-x))+C`