What is the integral of (x^2+x)^7(2x+1) ?

The integral of (x^2+x)^7(2x+1) is ((x^2+x)^8)/8+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+x)^7(2x+1)

\int (x^{2} + x)^{7} (2 x + 1) d x

\frac{( x ^{2} + x ) ^{8}}{8} + C

Solution steps

\int (x^{2} + x)^{7} (2 x + 1) d x

Apply u-substitution

: \int u^{7} d u

= \int u^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= \frac{u ^{8}}{8}

Substitute back

u = x^{2} + x

= \frac{( x ^{2} + x ) ^{8}}{8}

Add a constant to the solution

= \frac{( x ^{2} + x ) ^{8}}{8} + C

\int_{0}^{1} e^{x} - e^{- x} d x

f (x) = e^{x} (9 - 3 x + 2 x^{2}), at x = 0

\int \frac{e ^{4 x}}{1 + e ^{x}} d x

\int t (1 - ln^{2} (t)) d t

\int (- 2 sin (x) + 4 cos (x)) d x

What is the integral of (x^2+x)^7(2x+1) ?
The integral of (x^2+x)^7(2x+1) is ((x^2+x)^8)/8+C