Answer to Question #117592 in Calculus for Tanisha

Question #117592
Integrate the following functions wrt x
I) 1/(2x+1)^3/2
ii) sin(2x+3)
iii)cosec(4x)
iv)1/√(1-9x^2)
v)1/(1+4x)
1
Expert's answer
2020-05-25T18:51:46-0400

*all basic integrals you can find here https://math2.org/math/integrals/tableof.htm*1(2x+1)32dx=12(2x+1)32d(2x+1)=12x+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

sin(2x+3)dx=sin(2x+3)2d(2x+3)=cos(2x+3)+C\int sin(2x+3)dx = \int \frac{sin(2x+3)}{2}d(2x+3) = -cos(2x+3) + C

cosec(4x)dx=cosec(4x)4d(4x)=lncosec(4x)+cot(4x)4+C\int cosec(4x)dx = \int \frac{cosec(4x)}{4}d(4x) =-\frac{ln|cosec(4x)+cot(4x)|}{4} + C

119x2dx=1319x2d(3x)=sin13x3+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

11+4xdx=14(1+4x)d(4x)=ln(1+4x)4+C\int \frac{1}{1+4x}dx = \int \frac{1}{4(1+4x)}d(4x) = \frac{ln(1+4x)}{4} + C


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