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

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

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integral of 1/((x+1)sqrt(x^2+1))

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} (\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2})

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/((x+1)sqrt(x^2+1)) ?
The integral of 1/((x+1)sqrt(x^2+1)) is