What is the integral of (x^2+1)(4x-3) ?

The integral of (x^2+1)(4x-3) is (4x^4)/3-(x^4)/3-x^3+2x^2-3x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+1)(4x-3)

\int (x^{2} + 1) (4 x - 3) d x

\frac{4 x ^{4}}{3} - \frac{x ^{4}}{3} - x^{3} + 2 x^{2} - 3 x + C

Solution steps

\int (x^{2} + 1) (4 x - 3) d x

Apply Integration By Parts

: \frac{4 x ^{4}}{3} + 4 x^{2} - x^{3} - 3 x - \int 4 (\frac{x ^{3}}{3} + x) d x

= \frac{4 x ^{4}}{3} + 4 x^{2} - x^{3} - 3 x - \int 4 (\frac{x ^{3}}{3} + x) d x

\int 4 (\frac{x ^{3}}{3} + x) d x = 4 (\frac{x ^{4}}{12} + \frac{x ^{2}}{2})

= \frac{4 x ^{4}}{3} + 4 x^{2} - x^{3} - 3 x - 4 (\frac{x ^{4}}{12} + \frac{x ^{2}}{2})

Simplify

\frac{4 x ^{4}}{3} + 4 x^{2} - x^{3} - 3 x - 4 (\frac{x ^{4}}{12} + \frac{x ^{2}}{2}) : \frac{4 x ^{4}}{3} - \frac{x ^{4}}{3} - x^{3} + 2 x^{2} - 3 x

= \frac{4 x ^{4}}{3} - \frac{x ^{4}}{3} - x^{3} + 2 x^{2} - 3 x

Add a constant to the solution

= \frac{4 x ^{4}}{3} - \frac{x ^{4}}{3} - x^{3} + 2 x^{2} - 3 x + C

\int \frac{4 x ^{3}}{1 - x ^{4}} d x

\int x^{5} 6 x^{6} + 2 d x

\int (4 x^{2} + 3)^{2} d x

x^{2} - x, 2 x - 2 x^{2}

x \to 4 lim (\frac{1}{x + 2})

What is the integral of (x^2+1)(4x-3) ?
The integral of (x^2+1)(4x-3) is (4x^4)/3-(x^4)/3-x^3+2x^2-3x+C