Question

Integration of 5 x cube plus 2 x raised to minus 5 minus 7x plus one by root x plus 5 by x​

Answer 1

Answer:

$\bold\red{\frac{5 {x}^{4} }{4} - \frac{1}{2 {x}^{4} } - \frac{7 {x}^{2} }{2} + 2 \sqrt{x} + 5 \: ln( |x| ) + c}$

Step-by-step explanation:

Given,

$\int(5 {x}^{3} + 2 {x}^{ - 5} - 7x + \frac{1}{ \sqrt{x} } + \frac{5}{x} )dx \\ \\ = 5\int {x}^{3} dx + 2\int {x}^{ - 5} dx - 7\int \: xdx + \int {x}^{ \frac{ - 1}{2} } dx + 5\int \frac{1}{x} dx \\ \\ = 5 \times \frac{ {x}^{4} }{4} + 2 \times \frac{ {x}^{ - 4} }{( - 4)} - 7 \times \frac{ {x}^{2} }{2} + \frac{ {x}^{ \frac{1}{2} } }{( \frac{1}{2}) } + 5 \: ln( |x| ) + c \\ \\ = \frac{5 {x}^{4} }{4} - \frac{1}{2 {x}^{4} } - \frac{7 {x}^{2} }{2} + 2 \sqrt{x} + 5 \: ln( |x| ) + c$

Hence,

We get the integration value as

$\bold{\frac{5 {x}^{4} }{4} - \frac{1}{2 {x}^{4} } - \frac{7 {x}^{2} }{2} + 2 \sqrt{x} + 5 \: ln( |x| ) + c}$

Answer 2

Answer:

(5/4)x⁴-(1/2)x⁻⁴-(7/2)x²+2√x+5logx+c

Step-by-step explanation:

formulee used---->

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1)∫xⁿdx=(xⁿ⁺¹/n+1)+c, where n≠-1

2)∫(1/x)dx=logx+c

Now returning to original question

Let

I= ∫{5x³+2x⁻⁵-7x+(1/√x)+(5/x)}dx

We do intregation one by one

∫5x³dx=5 (x³⁺¹/3+1)=(5/4)x⁴

∫2x⁻⁵dx=2(x⁻⁵⁺¹/-5+1)=(2/-4)x⁻⁴=(-1/2)x⁻⁴

∫7xdx =7(x¹+¹/1+1)=(7/2)x²

∫(1/√x)dx=∫x⁻¹/²dx=(x⁻¹/²+¹)/(-1/2)+1

=2x¹/²

∫(5/x)dx=5logx

Now

I=(5/4)x⁴-(1/2)x⁻⁴-(7/2)x²+2x¹/²+5logx+c

Additional information---->

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1)∫eˣdx=eˣ+c

2)∫aˣdx=(aˣ/loga)+c

3)∫sinxdx=-cosx+c

4)∫cosxdx=sinx+c

5)∫sec²xdx=tanx+c

6)∫secx tanx dx=secx+c

7)∫cosec²xdx=-cotx+c

8)∫cosecx cotx dx=-cosecx+c