What is the integral of 1/(xsqrt(x^2-2x+5)) ?

The integral of 1/(xsqrt(x^2-2x+5)) is

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integral of 1/(xsqrt(x^2-2x+5))

\int \frac{1}{x x ^{2} - 2 x + 5} d x

\frac{1}{5} ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1 + C

Solution steps

\int \frac{1}{x x ^{2} - 2 x + 5} d x

Complete the square

x^{2} - 2 x + 5 : (x - 1)^{2} + 4

= \int \frac{1}{x ( x - 1 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{( u + 1 ) u ^{2} + 4} d u

= \int \frac{1}{( u + 1 ) u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2 tan ( v ) + 1} d v

= \int \frac{sec ( v )}{2 tan ( v ) + 1} d v

Apply u-substitution

: \int \frac{2}{4 w + 1 - w ^{2}} d w

= \int \frac{2}{4 w + 1 - w ^{2}} d w

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 w + 1 - w ^{2}} d w

Factor

4 w + 1 - w^{2} : - (- 4 w - 1 + w^{2})

= 2 \cdot \int \frac{1}{- ( - 4 w - 1 + w ^{2} )} d w

Take the constant out:

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

= 2 (- \int \frac{1}{- 4 w - 1 + w ^{2}} d w)

Complete the square

- 4 w - 1 + w^{2} : (w - 2)^{2} - 5

= 2 (- \int \frac{1}{( w - 2 ) ^{2} - 5} d w)

Apply u-substitution

: \int \frac{1}{t ^{2} - 5} d t

= 2 (- \int \frac{1}{t ^{2} - 5} d t)

Apply Integral Substitution

= 2 (- \int \frac{1}{5 ( r ^{2} - 1 )} d r)

Take the constant out:

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

= 2 (- \frac{1}{5} \cdot \int \frac{1}{r ^{2} - 1} d r)

Factor

r^{2} - 1 : - (- r^{2} + 1)

= 2 (- \frac{1}{5} \cdot \int \frac{1}{- ( - r ^{2} + 1 )} d r)

Take the constant out:

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

= 2 (- \frac{1}{5} (- \int \frac{1}{- r ^{2} + 1} d r))

Use the common integral:

\int \frac{1}{- r ^{2} + 1} d r = \frac{ln ∣ r + 1 ∣}{2} - \frac{ln ∣ r - 1 ∣}{2}

= 2 (- \frac{1}{5} (- (\frac{ln ∣ r + 1 ∣}{2} - \frac{ln ∣ r - 1 ∣}{2})))

Substitute back

= 2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1}{2}

Simplify

2 - \frac{1}{5} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1}{2} - \frac{ln \frac{t a n ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1}{2} : \frac{1}{5} ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1

= \frac{1}{5} ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1

Add a constant to the solution

= \frac{1}{5} ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} + 1 - ln \frac{tan ( \frac{a r c t a n ( \frac{1}{2} ( x - 1 ) )}{2} ) - 2}{5} - 1 + C

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What is the integral of 1/(xsqrt(x^2-2x+5)) ?
The integral of 1/(xsqrt(x^2-2x+5)) is