What is the integral of e^{3x}(1+e^{3x})^7 ?

The integral of e^{3x}(1+e^{3x})^7 is

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integral of e^{3x}(1+e^{3x})^7

\int e^{3 x} (1 + e^{3 x})^{7} d x

\frac{1}{3} e^{3 x} + \frac{7}{6} e^{6 x} + \frac{7}{3} e^{9 x} + \frac{35}{12} e^{12 x} + \frac{7}{3} e^{15 x} + \frac{7}{6} e^{18 x} + \frac{1}{3} e^{21 x} + \frac{1}{24} e^{24 x} + C

Solution steps

\int e^{3 x} (1 + e^{3 x})^{7} d x

Expand

e^{3 x} (1 + e^{3 x})^{7} : e^{3 x} + 7 e^{6 x} + 21 e^{9 x} + 35 e^{12 x} + 35 e^{15 x} + 21 e^{18 x} + 7 e^{21 x} + e^{24 x}

= \int e^{3 x} + 7 e^{6 x} + 21 e^{9 x} + 35 e^{12 x} + 35 e^{15 x} + 21 e^{18 x} + 7 e^{21 x} + e^{24 x} 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 e^{3 x} d x + \int 7 e^{6 x} d x + \int 21 e^{9 x} d x + \int 35 e^{12 x} d x + \int 35 e^{15 x} d x + \int 21 e^{18 x} d x + \int 7 e^{21 x} d x + \int e^{24 x} d x

\int e^{3 x} d x = \frac{1}{3} e^{3 x}

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

\int 21 e^{9 x} d x = \frac{7}{3} e^{9 x}

\int 35 e^{12 x} d x = \frac{35}{12} e^{12 x}

\int 35 e^{15 x} d x = \frac{7}{3} e^{15 x}

\int 21 e^{18 x} d x = \frac{7}{6} e^{18 x}

\int 7 e^{21 x} d x = \frac{1}{3} e^{21 x}

\int e^{24 x} d x = \frac{1}{24} e^{24 x}

= \frac{1}{3} e^{3 x} + \frac{7}{6} e^{6 x} + \frac{7}{3} e^{9 x} + \frac{35}{12} e^{12 x} + \frac{7}{3} e^{15 x} + \frac{7}{6} e^{18 x} + \frac{1}{3} e^{21 x} + \frac{1}{24} e^{24 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} + \frac{7}{6} e^{6 x} + \frac{7}{3} e^{9 x} + \frac{35}{12} e^{12 x} + \frac{7}{3} e^{15 x} + \frac{7}{6} e^{18 x} + \frac{1}{3} e^{21 x} + \frac{1}{24} e^{24 x} + C

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

\int_{- 6}^{5} x^{2} - 3 x - 4 d x

cos^{2} (x)

\int \frac{x ^{2}}{( x ^{2} + 16 ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial x} (\frac{6 x}{y ^{2}})

What is the integral of e^{3x}(1+e^{3x})^7 ?
The integral of e^{3x}(1+e^{3x})^7 is