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

The integral from 0 to 1 of 1/(2t^2-2t+1) is pi/2

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integral from 0 to 1 of 1/(2t^2-2t+1)

\int_{0}^{1} \frac{1}{2 t ^{2} - 2 t + 1} d t

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{1} \frac{1}{2 t ^{2} - 2 t + 1} d t

Complete the square

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

= \int_{0}^{1} \frac{1}{2 ( t - \frac{1}{2} ) ^{2} + \frac{1}{2}} d t

Apply u-substitution

: \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{2}{4 u ^{2} + 1} d u

= \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{2}{4 u ^{2} + 1} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{- \frac{1}{2}}^{\frac{1}{2}} \frac{1}{4 u ^{2} + 1} d u

Apply Integral Substitution

= 2 \cdot \int_{- 1}^{1} \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int_{- 1}^{1} \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{2} [arctan (v)]_{- 1}^{1}

Simplify

2 \cdot \frac{1}{2} [arctan (v)]_{- 1}^{1} : [arctan (v)]_{- 1}^{1}

= [arctan (v)]_{- 1}^{1}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

\frac{1}{1 - x ^{2}}, 0

\frac{d}{d x} (- \frac{x}{5})

x \to 2 lim (3 x^{3} - 5 x)

\int_{100}^{120} \frac{600}{x} d x

f (x) = 4 x^{2} - x^{3}, at x = 1

What is the integral from 0 to 1 of 1/(2t^2-2t+1) ?
The integral from 0 to 1 of 1/(2t^2-2t+1) is pi/2