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
Evaluate each of the following integral:
Solve each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
Find: `int logx/(1 + log x)^2 dx`
