# Integration Questions

## What is the integration of $$e^x$$ ?

Solution : The integration of $$e^x$$ with respect to x is $$e^x$$ + C. Since $$d\over dx$$ $$e^x$$ = $$e^x$$ dx On integrating both sides, we get $$\int$$ $$e^x$$ dx = $$e^x$$ Hence, the integration of $$e^x$$ is $$e^x$$ + C

## Integrate $$x^2 + x – 1\over x^2 – 1$$ with respect to x.

Solution : $$\int$$ $$x^2 + x – 1\over x^2 – 1$$ dx = $$\int$$ ($$x^2 – 1\over x^2 – 1$$ + $$x\over x^2 – 1$$)dx = $$\int$$ 1 dx + $$\int$$ $$x\over x^2 – 1$$) dx Let $$x^2 – 1$$ = t  $$\implies$$ 2x dx = dt = x + $$\int$$ $$dt\over 2t$$ = x …

## What is the integration of log cos x dx ?

Solution : We have, I = $$\int$$ log cos x dx By using integraton by parts, I = $$\int$$ 1.log cos x dx Taking log cos x as first function and 1 as second function. Then, I = log cos x $$\int$$ 1 dx – $$\int$$ { $${d\over dx}$$ (log cos x) $$\int$$ 1 dx …

## What is the integration of log 1/x ?

Solution :  We have, I = $$\int$$ $$log {1\over x}$$ dx I = $$\int$$ $$log 1 – log x$$ dx = $$\int$$ (-log x) dx By using integration by parts formula, Let I = -($$\int$$ log x .1) dx where log x is the first function and 1 is the second function according to ilate …

## What is the integration of 1/x log x ?

Solution : We have, I = $$\int$$ $$1\over x log x$$ dx Put log x = t $$\implies$$ $$1\over x$$ dx = dt I = $$\int$$  $$1\over t$$ dt I = log | t | + C I = log |log x| + C Hence, the integration of $$1\over x log x$$ is log (log …

Solution : We have, I = $$\int$$ x log x dx By using integration by parts, And taking log x as first function and x as second function. Then, I = log x { $$\int$$ x dx } – $$\int$$ { $${d\over dx}(log x) \times \int x dx$$ } dx I = (log x) $$x^2\over … ## What is the integration of \((log x)^2$$ dx ?

Solution : We have, I = $$(log x)^2$$ . 1 dx, Then , where $$(log x)^2$$ is the first function and 1 is the second function according to ilate rule, I = $$(log x)^2$$ { $$\int$$ 1 dx} – $$\int$$ {$$d\over dx$$ $$(log x)^2$$ . $$\int$$ 1 dx } dx I = $$(log x)^2$$ x …

## What is the integration of sec inverse root x ?

Solution : We have, I = $$sec^{-1}\sqrt{x}$$ dx Let x = $$sec^2t$$ dx = $$2sec^2 t tan t$$ dt I = t.$$2sec^2 t tan t$$ dt u = t  and v = $$tan^2 t$$ I = $$\int$$ u.dv = u.v – $$\int$$ v.du = $$t.tan^2 t$$ – $$\int$$ $$tan^2 t$$ dt I = $$t.tan^2 t$$ …

## What is the integration of cos inverse root x ?

Solution : We have, I = $$cos^{-1}\sqrt{x}$$ . 1 dx By Applying integration by parts, Taking $$cos^{-1}\sqrt{x}$$ as first function and 1 as second function. Then I = $$cos^{-1}\sqrt{x}$$ $$\int$$ 1 dx – $$\int$$ {$$d\over dx$$$$cos^{-1}\sqrt{x}$$ $$\int$$ 1 dx } dx I = x$$cos^{-1}\sqrt{x}$$ – $$\int$$ $$-1\over 2\sqrt{(1-x)}\sqrt{x}$$ . x dx I = x$$cos^{-1}\sqrt{x}$$ – …

## What is the integration of x cos inverse x ?

Solution : We have, I = $$\int$$  $$x cos^{-1} x$$ dx By using integration by parts formula, I = $$cos^{-1} x$$ $$x^2\over 2$$ – $$\int$$ $$-1\over \sqrt{1 – x^2}$$ $$\times$$ $$x^2\over 2$$ dx I =  $$x^2\over 2$$ $$cos^{-1} x$$ – $$1\over 2$$ $$\int$$ $$-x^2\over \sqrt{1 – x^2}$$ dx = $$x^2\over 2$$ $$cos^{-1} x$$ – \(1\over …