What is the integral of ((x^2+1)^3)/x ?

The integral of ((x^2+1)^3)/x is (x^2+1)^3-(5x^6)/6-(9x^4)/4-(3x^2)/2+ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int \frac{( x ^{2} + 1 ) ^{3}}{x} d x

(x^{2} + 1)^{3} - \frac{5 x ^{6}}{6} - \frac{9 x ^{4}}{4} - \frac{3 x ^{2}}{2} + ln ∣ x ∣ + C

Solution steps

\int \frac{( x ^{2} + 1 ) ^{3}}{x} d x

Apply Integration By Parts

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

\int \frac{6 x ^{2} ( x ^{2} + 1 ) ^{2} - ( x ^{2} + 1 ) ^{3}}{x} d x = \frac{5 x ^{6}}{6} + \frac{9 x ^{4}}{4} + \frac{3 x ^{2}}{2} - ln ∣ x ∣

= (x^{2} + 1)^{3} - (\frac{5 x ^{6}}{6} + \frac{9 x ^{4}}{4} + \frac{3 x ^{2}}{2} - ln ∣ x ∣)

Simplify

= (x^{2} + 1)^{3} - \frac{5 x ^{6}}{6} - \frac{9 x ^{4}}{4} - \frac{3 x ^{2}}{2} + ln ∣ x ∣

Add a constant to the solution

= (x^{2} + 1)^{3} - \frac{5 x ^{6}}{6} - \frac{9 x ^{4}}{4} - \frac{3 x ^{2}}{2} + ln ∣ x ∣ + C

What is the integral of ((x^2+1)^3)/x ?
The integral of ((x^2+1)^3)/x is (x^2+1)^3-(5x^6)/6-(9x^4)/4-(3x^2)/2+ln|x|+C