What is the integral of (3x^5-27x)(5x^4-9) ?

The integral of (3x^5-27x)(5x^4-9) is (3x^{10})/2-27x^6+(243x^2)/2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^5-27x)(5x^4-9)

\int (3 x^{5} - 27 x) (5 x^{4} - 9) d x

\frac{3 x ^{10}}{2} - 27 x^{6} + \frac{243 x ^{2}}{2} + C

Solution steps

\int (3 x^{5} - 27 x) (5 x^{4} - 9) d x

Expand

(3 x^{5} - 27 x) (5 x^{4} - 9) : 15 x^{9} - 162 x^{5} + 243 x

= \int 15 x^{9} - 162 x^{5} + 243 x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 15 x^{9} d x - \int 162 x^{5} d x + \int 243 x d x

\int 15 x^{9} d x = \frac{3 x ^{10}}{2}

\int 162 x^{5} d x = 27 x^{6}

\int 243 x d x = \frac{243 x ^{2}}{2}

= \frac{3 x ^{10}}{2} - 27 x^{6} + \frac{243 x ^{2}}{2}

Add a constant to the solution

= \frac{3 x ^{10}}{2} - 27 x^{6} + \frac{243 x ^{2}}{2} + C

\int \frac{4 x ^{2} - 4 x + 6}{x ^{3} - x ^{2} - 6 x} d x

\int (t + 1) (t + 1) d t

\int \frac{1}{sin ^{2} ( 4 θ )} d θ

\int \frac{( sec ^{2} ( ln ( x ) ) )}{x} d x

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

What is the integral of (3x^5-27x)(5x^4-9) ?
The integral of (3x^5-27x)(5x^4-9) is (3x^{10})/2-27x^6+(243x^2)/2+C