Question

Integrate(x^2-cos x +1/x)dx

Answer 1

GIVEN :–

A function x² - cos(x) + (1/x).

TO FIND :–

Intigration of given function.

SOLUTION :–

I = ∫ [x² - cos(x) + (1/x)]dx

• We know that –

(1) ∫(xⁿ)dx = [x^(n+1)] / (n+1)

(2) ∫cos(x)dx = sin(x)

(3) ∫(1/x) dx = log(x)

So that –

I = x³/3 - sin(x) + log(x) + c

Important formula of integration :–

(1) ∫sin(x).dx = - cos(x) + c

(2) ∫cos(x).dx = sin(x) + c

(3) ∫tan(x).dx = log|sec(x)| +c

(4) ∫cot(x).dx = log|sin(x)| + c

(5) ∫sec(x).dx = log|sec(x) + tan(x)| + c

(6) ∫cosec(x).dx = log|cosec(x) - cot(x)| + c

Answer 2

Answer:

$\displaystyle{\frac{x^3}{3} - sinx + log_ex + c}$

Explanation:

$\displaystyle{\int{\bigg(x^2 - cosx + \frac{1}{x}\bigg)dx}}$

$\displaystyle{= \int {x^2\:dx} - \int {cosx\:dx} + \int {\frac{1}{x}\:dx}}$

$\displaystyle{= \frac{x^{2+3}}{2+1} - sinx + log_ex + c}$

$\displaystyle{= \frac{x^3}{3} - sinx + log_ex + c}$

Basic formulae of integration:

$\displaystyle{\sf{1. \; \int{ x^ndx} = \frac{x^{n+1}}{n+1} + c \quad (n \neq -1)}}$

$\displaystyle{\sf{2. \; \int {sinxdx} = -cosx + c }}$

$\displaystyle{\sf{3. \; \int {cosxdx} = sinx + c }}$

$\displaystyle{\sf{4. \; \int {\frac{1}{x}dx} = log_ex + c }}$

$\displaystyle{\sf{5. \; \int{e^xdx} = e^x + c }}$