Advertisements
Advertisements
प्रश्न
Evaluate: ∫ (log x)2 dx
उत्तर
Let I = ∫ (log x)2 dx
I = ∫ (log x)2 . 1 dx
I = `(log x)^2 int 1. "dx" − int ["d"/"dx" (log x)^2 int 1. "dx"] "dx"`
I = `x(log x)^2 − int 2 log x. 1/cancelx. cancelx "dx"`
I = `x(log x)^2 − 2 int log x. 1 "dx"`
I = `x(log x)^2 − 2[log x int 1. "dx" − int {"d"/"dx" (log x) int 1. "dx"}]`dx
I = `x(log x)^2 − 2[(log x)x − int 1/cancelx. cancelx. "dx"]`
I = `x(log x)^2 − 2[xlog x − int 1. "dx"]`
I = `x(log x)^2 − 2(x log x − x) + c`
∴ I = x(log x)2 − 2x log x + 2x + c
APPEARS IN
संबंधित प्रश्न
Integrate the function in x log x.
Evaluate the following : `int cos sqrt(x).dx`
Integrate the following functions w.r.t. x : `log(1 + x)^((1 + x)`
Integrate the following with respect to the respective variable : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`
Evaluate the following.
`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx
`int ("x" + 1/"x")^3 "dx"` = ______
Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx
Choose the correct alternative:
`int ("d"x)/((x - 8)(x + 7))` =
`int [(log x - 1)/(1 + (log x)^2)]^2`dx = ?
∫ log x · (log x + 2) dx = ?
Find `int_0^1 x(tan^-1x) "d"x`
`intsqrt(1+x) dx` = ______
Evaluate:
`intcos^-1(sqrt(x))dx`
Evaluate:
`inte^x sinx dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following:
`intx^3e^(x^2)dx`
Evaluate the following.
`intx^3e^(x^2) dx`
Evaluate:
`int x^2 cos x dx`