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 \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
