What is the integral of (4sqrt(x))/(sqrt(1+x^3)) ?

The integral of (4sqrt(x))/(sqrt(1+x^3)) is 8/3 ln|sqrt(x^3+1)+x^{3/2}|+C

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integral of (4sqrt(x))/(sqrt(1+x^3))

\int \frac{4 x}{1 + x ^{3}} d x

\frac{8}{3} ln x^{3} + 1 + x^{\frac{3}{2}} + C

Solution steps

\int \frac{4 x}{1 + x ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{1 + x ^{3}} d x

Apply u-substitution

= 4 \cdot \int \frac{2 u ^{2}}{1 + u ^{6}} d u

Take the constant out:

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

= 4 \cdot 2 \cdot \int \frac{u ^{2}}{1 + u ^{6}} d u

Apply u-substitution

= 4 \cdot 2 \cdot \int \frac{1}{3 1 + v ^{2}} d v

Take the constant out:

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

= 4 \cdot 2 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= 4 \cdot 2 \cdot \frac{1}{3} ln v^{2} + 1 + v

Substitute back

= 4 \cdot 2 \cdot \frac{1}{3} ln ((x)^{3})^{2} + 1 + (x)^{3}

Simplify

4 \cdot 2 \cdot \frac{1}{3} ln ((x)^{3})^{2} + 1 + (x)^{3} : \frac{8}{3} ln x^{3} + 1 + x^{\frac{3}{2}}

= \frac{8}{3} ln x^{3} + 1 + x^{\frac{3}{2}}

Add a constant to the solution

= \frac{8}{3} ln x^{3} + 1 + x^{\frac{3}{2}} + C

What is the integral of (4sqrt(x))/(sqrt(1+x^3)) ?
The integral of (4sqrt(x))/(sqrt(1+x^3)) is 8/3 ln|sqrt(x^3+1)+x^{3/2}|+C