What is the integral of (2x^2-3)^5 ?

The integral of (2x^2-3)^5 is (32x^{11}}{11}-\frac{80x^9)/3+(720x^7)/7-216x^5+270x^3-243x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^2-3)^5

\int (2 x^{2} - 3)^{5} d x

\frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x + C

Solution steps

\int (2 x^{2} - 3)^{5} d x

Expand

(2 x^{2} - 3)^{5} : 32 x^{10} - 240 x^{8} + 720 x^{6} - 1080 x^{4} + 810 x^{2} - 243

= \int 32 x^{10} - 240 x^{8} + 720 x^{6} - 1080 x^{4} + 810 x^{2} - 243 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 32 x^{10} d x - \int 240 x^{8} d x + \int 720 x^{6} d x - \int 1080 x^{4} d x + \int 810 x^{2} d x - \int 243 d x

\int 32 x^{10} d x = \frac{32 x ^{11}}{11}

\int 240 x^{8} d x = \frac{80 x ^{9}}{3}

\int 720 x^{6} d x = \frac{720 x ^{7}}{7}

\int 1080 x^{4} d x = 216 x^{5}

\int 810 x^{2} d x = 270 x^{3}

\int 243 d x = 243 x

= \frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x

Add a constant to the solution

= \frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x + C

\int e^{7 x} cos (4 x) d x

y^{^{'}} = (x + y - 3)^{2}

\int - x tan (x) d x

x \to 1 lim (2 x^{2} - 5 x - 1)

6 e^{- 5 t} + e^{3 t} + 5 t^{3} - 9

What is the integral of (2x^2-3)^5 ?
The integral of (2x^2-3)^5 is (32x^{11}}{11}-\frac{80x^9)/3+(720x^7)/7-216x^5+270x^3-243x+C