Question

Integration of (x^2+1/x^2)^3

Answer 1

Answer:

integral(x^2 + 1/x^2)^3 dx = x^7/7 - 1/(5 x^5) + x^3 - 3/x + constant

Answer 2

Your Answer:-

To find:-

• Evaluate  $\tt \int( x^2 +\dfrac{1}{x^2})^3 d_x$.

Solution:-

$\tt \int( x^2 +\dfrac{1}{x^2})^3{d}x \\\\ \\\tt = \int [x^6 +\dfrac{1}{x^6}+3(x^2)(\dfrac{1}{x^2})(x^2+\dfrac{1}{x^2})]{d}x \\\\ \\\because (a+b)^3=a^3+b^3+3ab(a+b)$

$\tt= \int [x^6 +\dfrac{1}{x^6}+3(\cancel{x^2})(\dfrac{1}{\cancel{x^2}})(x^2+\dfrac{1}{x^2})]{d}x \\\\\\ \tt =\int [x^6 +\dfrac{1}{x^6}+(x^2+\dfrac{1}{x^2})]{d}x$

$\tt = \int x^6{d}x+\int \dfrac{1}{x^6}{d}x+3 \int x^2 {d}x+3\int \dfrac{1}{x^2}{d}x \\\\ \\\tt =\dfrac{x^7}{7}+ \dfrac{x^{-6+1}}{-6+1}+3[\dfrac{x^{2+1}}{2+1}]+3[\dfrac{x^{-2+1}}{-2+1}]+C \\\\ \\\tt = \dfrac{x^7}{7}+(-\dfrac{1}{5x^5})+x^3+(-\dfrac{3}{x})+C \\\\\\ \tt =\dfrac{x^7}{7}-\dfrac{1}{5x^5}+x^3-\dfrac{3}{x}+C$

Integral Rule Used:-

• Sum Rule: $\tt \int (f + g) {d}x = \int f {d}x + \int g {d}x$

Some Other Rules :-

• Difference Rule: $\tt \int (f - g) dx = \int f dx - \int g dx$

• Power Rule: If (n≠-1) then $\tt \int x^n dx = \dfrac{x^{n+1}}{n+1} + C$