*all basic integrals you can find here https://math2.org/math/integrals/tableof.htm *∫ 1 ( 2 x + 1 ) 3 2 d x = ∫ 1 2 ( 2 x + 1 ) 3 2 d ( 2 x + 1 ) = 1 2 x + 1 + C \int \frac{1}{(2x+1)^{\frac{3}{2}}}dx = \int \frac{1}{2(2x+1)^{\frac{3}{2}}}d(2x+1) = \frac{1}{\sqrt{2x+1}} + C ∫ ( 2 x + 1 ) 2 3 1 d x = ∫ 2 ( 2 x + 1 ) 2 3 1 d ( 2 x + 1 ) = 2 x + 1 1 + C
∫ s i n ( 2 x + 3 ) d x = ∫ s i n ( 2 x + 3 ) 2 d ( 2 x + 3 ) = − c o s ( 2 x + 3 ) + C \int sin(2x+3)dx = \int \frac{sin(2x+3)}{2}d(2x+3) = -cos(2x+3) + C ∫ s in ( 2 x + 3 ) d x = ∫ 2 s in ( 2 x + 3 ) d ( 2 x + 3 ) = − cos ( 2 x + 3 ) + C
∫ c o s e c ( 4 x ) d x = ∫ c o s e c ( 4 x ) 4 d ( 4 x ) = − l n ∣ c o s e c ( 4 x ) + c o t ( 4 x ) ∣ 4 + C \int cosec(4x)dx = \int \frac{cosec(4x)}{4}d(4x) =-\frac{ln|cosec(4x)+cot(4x)|}{4} + C ∫ cosec ( 4 x ) d x = ∫ 4 cosec ( 4 x ) d ( 4 x ) = − 4 l n ∣ cosec ( 4 x ) + co t ( 4 x ) ∣ + C
∫ 1 1 − 9 x 2 d x = ∫ 1 3 1 − 9 x 2 d ( 3 x ) = s i n − 1 3 x 3 + C \int \frac{1}{\sqrt{1-9x^2}}dx = \int \frac{1}{3\sqrt{1-9x^2}}d(3x) = \frac{sin^{-1}3x}{3} + C ∫ 1 − 9 x 2 1 d x = ∫ 3 1 − 9 x 2 1 d ( 3 x ) = 3 s i n − 1 3 x + C
∫ 1 1 + 4 x d x = ∫ 1 4 ( 1 + 4 x ) d ( 4 x ) = l n ( 1 + 4 x ) 4 + C \int \frac{1}{1+4x}dx = \int \frac{1}{4(1+4x)}d(4x) = \frac{ln(1+4x)}{4} + C ∫ 1 + 4 x 1 d x = ∫ 4 ( 1 + 4 x ) 1 d ( 4 x ) = 4 l n ( 1 + 4 x ) + C
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