What is the integral from 0 to 1 of (x^3)/(x^2+2x+1) ?

The integral from 0 to 1 of (x^3)/(x^2+2x+1) is 3ln(2)-2

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integral from 0 to 1 of (x^3)/(x^2+2x+1)

\int_{0}^{1} \frac{x ^{3}}{x ^{2} + 2 x + 1} d x

3 ln (2) - 2

Decimal

0.07944 \dots

Solution steps

\int_{0}^{1} \frac{x ^{3}}{x ^{2} + 2 x + 1} d x

Complete the square

x^{2} + 2 x + 1 : (x + 1)^{2}

= \int_{0}^{1} \frac{x ^{3}}{( x + 1 ) ^{2}} d x

Apply u-substitution

: \int_{1}^{2} \frac{( u - 1 ) ^{3}}{u ^{2}} d u

= \int_{1}^{2} \frac{( u - 1 ) ^{3}}{u ^{2}} d u

Expand

\frac{( u - 1 ) ^{3}}{u ^{2}} : u - 3 + \frac{3}{u} - \frac{1}{u ^{2}}

= \int_{1}^{2} u - 3 + \frac{3}{u} - \frac{1}{u ^{2}} d u

Apply the Sum Rule:

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

= \int_{1}^{2} u d u - \int_{1}^{2} 3 d u + \int_{1}^{2} \frac{3}{u} d u - \int_{1}^{2} \frac{1}{u ^{2}} d u

\int_{1}^{2} u d u = \frac{3}{2}

\int_{1}^{2} 3 d u = 3

\int_{1}^{2} \frac{3}{u} d u = 3 ln (2)

\int_{1}^{2} \frac{1}{u ^{2}} d u = \frac{1}{2}

= \frac{3}{2} - 3 + 3 ln (2) - \frac{1}{2}

Simplify

\frac{3}{2} - 3 + 3 ln (2) - \frac{1}{2} : 3 ln (2) - 2

= 3 ln (2) - 2

f (x) = - \frac{12}{x ^{5}}

x \to 3 lim (∣ 3 - x ∣)

\int e^{c o s (8 t)} sin (8 t) d t

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

\frac{\partial}{\partial x} (- 5 x^{2} + y^{2} + 3 z^{2} + 109)

What is the integral from 0 to 1 of (x^3)/(x^2+2x+1) ?
The integral from 0 to 1 of (x^3)/(x^2+2x+1) is 3ln(2)-2