Advertisements
Advertisements
Question
`int x^3/(x + 1)` is equal to ______.
Options
`x + x^2/2 + x^3/3 - log|1 - x| + "C"`
`x + x^2/2 - x^3/3 - log|1 - x| + "C"`
`x - x^2/2 - x^3/3 - log|1 + x| + "C"`
`x - x^2/2 + x^3/3 - log|1 + x| + "C"`
Advertisements
Solution
`int x^3/(x + 1)` is equal to `x - x^2/2 + x^3/3 - log|1 + x| + "C"`.
Explanation:
I = `int x^3/(x + 1)`
= `int (x^3 + 1 - 1)/(x + 1) "d"x`
= `int (x^3 + 1)/(x + 1) "d"x - int 1/(x + 1) "d"x`
= `int (x^2 - x + 1)"d"x - int 1/(x + 1) "d"x`
= `x^3/3 - x^2/2 + x - log|x + 1| + "C"`
APPEARS IN
RELATED QUESTIONS
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate each of the following integral:
Evaluate :
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^4 x dx\]
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
