What is the integral of 1/(xsqrt(1+x^n)) ?

The integral of 1/(xsqrt(1+x^n)) is -1/n (ln|sqrt(1+x^n)+1|-ln|sqrt(1+x^n)-1|)+C

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integral of 1/(xsqrt(1+x^n))

\int \frac{1}{x 1 + x ^{n}} d x

- \frac{1}{n} (ln 1 + x^{n} + 1 - ln 1 + x^{n} - 1) + C

Solution steps

\int \frac{1}{x 1 + x ^{n}} d x

Apply u-substitution

= \int \frac{1}{n ( u - 1 ) u} d u

Take the constant out:

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

= \frac{1}{n} \cdot \int \frac{1}{( u - 1 ) u} d u

Apply u-substitution

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

Take the constant out:

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

= \frac{1}{n} \cdot 2 (- (\frac{ln 1 + x ^{n} + 1}{2} - \frac{ln 1 + x ^{n} - 1}{2}))

Simplify

\frac{1}{n} \cdot 2 (- (\frac{ln 1 + x ^{n} + 1}{2} - \frac{ln 1 + x ^{n} - 1}{2})) : - \frac{1}{n} (ln 1 + x^{n} + 1 - ln 1 + x^{n} - 1)

= - \frac{1}{n} (ln 1 + x^{n} + 1 - ln 1 + x^{n} - 1)

Add a constant to the solution

= - \frac{1}{n} (ln 1 + x^{n} + 1 - ln 1 + x^{n} - 1) + C

x \to \infty lim ((a)^{x})

y^{^{'}} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1

d er i v a t i v e x^{3} b^{5} + y^{4} b + z^{6}

\int_{0}^{10} 100 e^{0.1 x} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 x e^{- x}

What is the integral of 1/(xsqrt(1+x^n)) ?
The integral of 1/(xsqrt(1+x^n)) is -1/n (ln|sqrt(1+x^n)+1|-ln|sqrt(1+x^n)-1|)+C