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

The integral of 1/(x*sqrt(x)+x^2) is 2(-ln|sqrt(x)|-1/(sqrt(x))+ln|sqrt(x)+1|)+C

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

\int \frac{1}{x \cdot x + x ^{2}} d x

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

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

Substitute back

u = x

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

Add a constant to the solution

= 2 (- ln x - \frac{1}{x} + ln x + 1) + C

\int_{0}^{e} x ln (x) d x

y^{^{'}} = (x - 2) (y - 2)

f (x) = - x + 5 x^{- \frac{1}{2}} - 3, (1, 1)

\frac{d y}{d x} = \frac{( x - 3 )}{y - 3}

\int_{0}^{\infty} \frac{1}{1 + x} d x

What is the integral of 1/(x*sqrt(x)+x^2) ?
The integral of 1/(x*sqrt(x)+x^2) is 2(-ln|sqrt(x)|-1/(sqrt(x))+ln|sqrt(x)+1|)+C