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

The integral of 1/((1+x)^{3/4)} is x/((1+x)^{3/4)}+(3x+4)/((1+x)^{3/4)}+C

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integral of 1/((1+x)^{3/4)}

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

\frac{x}{( 1 + x ) ^{\frac{3}{4}}} + \frac{3 x + 4}{( 1 + x ) ^{\frac{3}{4}}} + C

Solution steps

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

Apply Integration By Parts

: \frac{x}{( 1 + x ) ^{\frac{3}{4}}} - \int - \frac{3 x}{4 ( 1 + x ) ^{\frac{7}{4}}} d x

= \frac{x}{( 1 + x ) ^{\frac{3}{4}}} - \int - \frac{3 x}{4 ( 1 + x ) ^{\frac{7}{4}}} d x

\int - \frac{3 x}{4 ( 1 + x ) ^{\frac{7}{4}}} d x = - \frac{3 x + 4}{( 1 + x ) ^{\frac{3}{4}}}

= \frac{x}{( 1 + x ) ^{\frac{3}{4}}} - (- \frac{3 x + 4}{( 1 + x ) ^{\frac{3}{4}}})

Simplify

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

Add a constant to the solution

= \frac{x}{( 1 + x ) ^{\frac{3}{4}}} + \frac{3 x + 4}{( 1 + x ) ^{\frac{3}{4}}} + C

\int \frac{x ^{2} - x + 42}{x ^{3} + 7 x} d x

y = x^{\frac{1}{2}}

f (x) = 2 x + 5, at x = 2

n = 0 \sum \infty \frac{10 0 ^{n}}{n !}

\frac{1 - 2 s}{s ^{2} + 4 s + 5}

What is the integral of 1/((1+x)^{3/4)} ?
The integral of 1/((1+x)^{3/4)} is x/((1+x)^{3/4)}+(3x+4)/((1+x)^{3/4)}+C