What is the integral of (x^2+1)^4 ?

The integral of (x^2+1)^4 is (x^9)/9+(4x^7)/7+(6x^5)/5+(4x^3)/3+x+C

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integral of (x^2+1)^4

\int (x^{2} + 1)^{4} d x

\frac{x ^{9}}{9} + \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x + C

Solution steps

\int (x^{2} + 1)^{4} d x

Expand

(x^{2} + 1)^{4} : x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2} + 1

= \int x^{8} + 4 x^{6} + 6 x^{4} + 4 x^{2} + 1 d x

Apply the Sum Rule:

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

= \int x^{8} d x + \int 4 x^{6} d x + \int 6 x^{4} d x + \int 4 x^{2} d x + \int 1 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 1 d x = x

= \frac{x ^{9}}{9} + \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x

Add a constant to the solution

= \frac{x ^{9}}{9} + \frac{4 x ^{7}}{7} + \frac{6 x ^{5}}{5} + \frac{4 x ^{3}}{3} + x + C

{\frac{1}{s ^{3} + 5 s}}

\int e^{- 3 x} d x

\int \frac{41}{x ^{3} - 27} d x

\int \frac{x ^{3}}{3 x ^{2} + 1} d x

n = 1 \sum \infty \frac{1}{4 n + 3}

What is the integral of (x^2+1)^4 ?
The integral of (x^2+1)^4 is (x^9)/9+(4x^7)/7+(6x^5)/5+(4x^3)/3+x+C