Question

Evaluate the integral

Answer 1

The value of the integration is

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$, where C is an integrating constant.

We know that finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many tiny rectangles as possible, then add up their areas. The sum gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration. If a function can be integrated and its integral over the domain is finite with the given bounds, then the integration is definite.

Here, we have to determine the value of the given integral

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx$.

So,

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx = 7 \int {\frac{1-x^{3}-1 }{x^{3}+1} } \, dx$$= 7 \int [{1-\frac{1 }{x^{3}+1} } ]\, dx =7\int \, dx - 7 \int {\frac{1 }{x^{3}+1} } \, dx$ ...(1)

Now,

$\frac{1}{x^{3} +1} = \frac{1}{(x+1)(x^{2} -x+1)}$

We can write

$\frac{1}{(x+1)(x^{2} -x+1)}=\frac{A}{(x+1)} + \frac{Bx+C}{(x^{2} -x+1)}$

i.e. $1 = A(x^{2} -x+1)+(Bx+C)(x+1)$

i.e. $1 = Ax^{2} -Ax + A+Bx^{2} +Bx+Cx+C$

i.e. $1=(A+B)x^{2} +(-A+B+C)x +(A+C)$

Comparing both sides, we get

$A+B=0 \implies B = -A$ ...(2)

$-A+B+C = 0 \implies -A+(-A)+C=0 \implies -2A+C=0$ ...(3)

$A+C=1$ ...(4)

Subtracting (3) from (4),

$A+C+2A-C=1-0 \implies 3A=1 \implies A=\frac{1}{3}$

From (4), $C= 1-\frac{1}{3} =\frac{3-1}{3} =\frac{2}{3}$

From (2), $B =-\frac{1}{3}$

We get

$\frac{1}{x^{3}+1 } =\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{3} \frac{x-2}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-4}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1-3}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{3}{6} \frac{1}{x^{2} -x+1}$

$=\frac{1}{3} \frac{1}{(x+1)} - \frac{1}{6} \frac{2x-1}{x^{2} -x+1}+\frac{1}{2} \frac{1}{x^{2} -x+1}$

Integrate both sides,

$\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \int \frac{1}{(x+1)} \, dx - \frac{1}{6} \int \frac{2x-1}{x^{2} -x+1}\, dx+\frac{1}{2} \int \frac{1}{x^{2} -x+1}\, dx$

i.e. $\int \frac{1}{x^{3}+1 }\, dx =\frac{1}{3} \ln(x+1) - \frac{1}{6} ln (x^{2} -x+1)+\frac{\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })$

From (1),

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$

Therefore, the value of the integration is

$\int {\frac{-7x^{3} }{x^{3}+1} } \, dx =7x - \frac{7}{3} \ln(x+1) + \frac{7}{6} ln (x^{2} -x+1)-\frac{7\sqrt{3} }{3} tan^{-1} (\frac{2x-1}{\sqrt{3} })+C$, where C is an integrating constant.

Learn more about integrations here -

brainly.com/question/7541827

#SPJ10