What is the integral from 1 to 6 of 3/(x^2+2x+1) ?

The integral from 1 to 6 of 3/(x^2+2x+1) is 15/14

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integral from 1 to 6 of 3/(x^2+2x+1)

\int_{1}^{6} \frac{3}{x ^{2} + 2 x + 1} d x

\frac{15}{14}

Decimal

1.07142 \dots

Solution steps

\int_{1}^{6} \frac{3}{x ^{2} + 2 x + 1} d x

Take the constant out:

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

= 3 \cdot \int_{1}^{6} \frac{1}{x ^{2} + 2 x + 1} d x

Complete the square

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

= 3 \cdot \int_{1}^{6} \frac{1}{( x + 1 ) ^{2}} d x

Apply u-substitution

= 3 \cdot \int_{2}^{7} \frac{1}{u ^{2}} d u

Apply the Power Rule

= 3 [- \frac{1}{u}]_{2}^{7}

Compute the boundaries:

\frac{5}{14}

= 3 \cdot \frac{5}{14}

Simplify

= \frac{15}{14}

What is the integral from 1 to 6 of 3/(x^2+2x+1) ?
The integral from 1 to 6 of 3/(x^2+2x+1) is 15/14