Advertisements
Advertisements
Question
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Advertisements
Solution
`int((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2)) "d"x`
= `int 2"a"(x)^((-1)/2) "d"x - int "b"x^-2 "d"x + int 3"c" x^(2/3) "d"x`
= `4"a" sqrt(x) + "b"/x + (9"c"x^(5/3))/5 + "C"`.
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`Γ (9/2)`
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
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
