हिंदी

Evaluate the following. ∫13x2+8 dx

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प्रश्न

Evaluate the following.

`int 1/(sqrt(3"x"^2 + 8))` dx

योग
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उत्तर

Let I = `int 1/(sqrt(3"x"^2 + 8))` dx

`int 1/(sqrt((sqrt3"x")^2 + (sqrt8)^2))` dx

`= (log |sqrt3"x" + sqrt((sqrt3"x")^2 + (sqrt8)^2)|)/sqrt3` + c

∴ I = `1/sqrt3 log |sqrt3"x" + sqrt(3"x"^2 + 8)|` + c

Alternate method:

Let I = `"I" = int 1/sqrt(3"x"^2 + 8) "dx" = 1/sqrt3 int 1/(sqrt ("x"^2 + 8/3)` dx

`= 1/sqrt3 int 1/sqrt("x"^2 + ((2sqrt2)/sqrt3)^2)` dx

`= 1/sqrt3 log |"x" + sqrt("x"^2 + ((2sqrt2)/sqrt3)^2)| + "c"_1`

`= 1/sqrt3 log |"x" + sqrt("x"^2 + 8/3)| + "c"_1`

`= 1/sqrt3 log |(sqrt3"x" + sqrt(3"x"^2 + 8))/sqrt3| + "c"_1`

`= 1/sqrt3 log|sqrt3"x" + sqrt(3"x"^2 + 8)| - 1/sqrt3 log sqrt3 + "c"_1`

∴ I = `1/sqrt3 log |sqrt3"x" + sqrt(3"x"^2 + 8)|` + c

where c = `"c"_1 - 1/sqrt3 log sqrt3`

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Notes

The answer in the textbook is incorrect.

  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1.5: Integration - Q.4

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