4 - 6 Examples | Indefinite Integrals

$\displaystyle \int \sqrt{x^3 + 2} \,\, x^2 \, dx$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int (x^3 + 2)^{1/2} x^2 \, dx$
 

This is not of the form $\displaystyle \int u^n \, du$ because of the missing constant factor 3 in the integrand. Identifying $u = x^3 + 2$, $n = \frac{1}{2}$, then the differential $du = 3x^2 \, dx$. We must then insert 3 in the integrand and to compensate for it, we place the reciprocal 1/3 before the integral sign. This in effect multiplying by one does not affect the value of the function.
 

Let $u = x^3 + 2$, then $du = 3x^2 \, dx$
 

$\displaystyle \int \sqrt{x^3 + 2} \,\, x^2 \, dx$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int (x^3 + 2)^{1/2} x^2 \, dx$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \dfrac{1}{3}\int (x^3 + 2)^{1/2} (3x^2 \, dx)$

$\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{3} \, \dfrac{(x^3 + 2)^{3/2}}{3/2}$

$\,\,\,\,\,\,\,\,\,\, = \frac{2}{9} (x^3 + 2)^{3/2}$           answer

$\displaystyle \int \dfrac{(3x^2 + 1) \, dx}{\root 3\of {(2x^3 + 2x + 1)^2}}$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int \dfrac{(3x^2 + 1) \, dx}{(2x^3 + 2x + 1)^{2/3}}$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int (2x^3 + 2x + 1)^{-2/3}\,(3x^2 + 1) \, dx$
 

Let
u = 2x³ + 2x + 1
du = (6x² + 2) dx = 2(3x² + 1) dx
n = -2/3
 

$\displaystyle \int \dfrac{(3x^2 + 1) \, dx}{\root 3\of {(2x^3 + 2x + 1)^2}}$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \dfrac{1}{2}\int (2x^3 + 2x + 1)^{-2/3}\,[ \, 2(3x^2 + 1) \, ]\, dx$

$\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\,\dfrac{(2x^3 + 2x + 1)^{1/3}}{1/3} + C$

$\,\,\,\,\,\,\,\,\,\, = \frac{3}{2}\,(2x^3 + 2x + 1)^{1/3} + C$           answer

$\displaystyle \int (1 - 2x^2)^3 \, dx$

If we let $u = 1 - 2x^2$ and $n = 3$, then $du = -4x \, dx$. But there is no $x$ in the given integrand. It is easy to insert -4 in the integrand and offset this by placing -1/4 before the integral sign but nothing can be done about the missing factor $x$. We therefore expand $(1 - 2x^2)^3$ and integrate term by term.
 

$\displaystyle \int (1 - 2x^2)^3 \, dx$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int [ \, 1^3 - 3(1^2)(2x^2) + 3(1)(2x^2)^2 - (2x^2)^3 \, ] \, dx$

$\,\,\,\,\,\,\,\,\,\, = \displaystyle \int (1 - 6x^2 + 12x^4 - 8x^6) \, dx$

$\,\,\,\,\,\,\,\,\,\, = x - \dfrac{6x^3}{3} + \dfrac{12x^5}{5} - \dfrac{8x^7}{7} + C$

$\,\,\,\,\,\,\,\,\,\, = x - 2x^3 + \frac{12}{5}x^5 - \frac{8}{7}x^7 + C$           answer